IN  MEMORIAM 
FLORIAN  CAJORI 


Digitized  by  the  Internet  Archive 

in  2008  with  funding  from 

IVIicrosoft  Corporation 


http://www,archive.org/details/algebradesignedfOOstodrich 


AN 


ALGEBRA, 


faa^ 


GNED    FOK    THE    USE    OF 


HIGH  SCHOOLS,  ACADEMIES,  AND  COLLEGES. 


BY 


JOHN    F.UtODDAKD,  A.M., 

AUTnOK     OF     "STODDASD    S     AKITHMKTICAL     SKRIES,"     E*JO. 
AND 

PROF.  W.  D.  HENKLE, 

OF      OREFNMOCNT      C  O  L  L  E  (,  E .      INDIANA.  <i^ 


NEW     YORK: 
PUBLISHED    BY    SHELDON,   BLAKEAIAN    &   CO., 

No      115     NASSAU     STREET. 

1S57. 


^ 


C  ^  \  or  \ 


Entered,  according  to  Act  of  Congress,  in  the  year  1857,  by 

SHELDOX,  BLAKEMAX  &'C0., 

In  the  Clerk's  Office  of  the  District  Court  for  the  Southern  District  of  Xevf  York. 


BTBBKOTTPKD     BT 

THOMAS    B.    SMITH 
82  &84  ^pekman-Bt.,  N.  Y. 


7 


■Jhl^: 


/ 


V 


P  R  E  F  A  C  E 


Ji 


The  present  volume  is  designed  as  the  first  part  of 
a  complete  work  upon  the  science  of  Algebra.  Several 
chapters  written  for  this  volume  have  been  omitted  for 
want  of  room.  These  chapters  embrace  Indeterminate 
Analysis,  Permutations,  Combinations,  Calculus  of  Prob- 
abilities, Continued  Fractions,  and  about  fifty  pages 
on  the  General  Theory  of  Equations,  including  Sturm's 
Theorem,  Horner's  Method  of  Resolving  Numerical  Equa- 
tions of  alfy  Degrees,  and  Cardan's  Formula  for  Cubic 
Equations,  which  will  be  inserted  in  the  next  volume. 
The  present  volume,  however,  will  be  found  to  contain 
as  much  as  the  great  majority  of  students  in  our  High 
Schools  and  Academies  generally  study,  and  the  student 
who  thoroughly  masters  this  volume  will  have  acquired 
sufficient  algebraic  knowledge  to  enable  him  to  pursue 
the  remainder  of  the  usual  mathematical  course. 

The  arrangement  of  this  work  is,  in  many  respects, 
new.  The  plan  of  treating  every  subject  as  the  solu- 
tion of  a  general  problem,  or  as  the  demonstration  of 
a   theorem,   it  is  hoped  will   greatly  facilitate  rigid  class 


IV  P  EEF  ACE. 

% 

instruction.  Many  of  the -demonstrations  have  been  ren- 
dered so  clear  that  they  may  be  readily  comprehended 
by  students  of  ordinary  capacity  ;  others  which  are  of 
a  more  abstruse  character  will  require  close  application. 
The  student  should  earf-ly  endeavor  to  acquire  the  ability 
to  grasp  algebraic  principles  in  all  their  generality,  and 
hence  no  principle  has  been  given  with  a  mere  t71ustra^ 
tion,  which  is  too  often  considered  a  c^emonstration. 

The  attention  of  teachers  is  called  to  the  classifica- 
tion of  Algebraic  Symbols  in  Chap.  I. ;  the  explaiia- 
tion  of  Subtraction  and  Articles  (81),  (82),  (83),  (84), 
"(94),  (95),  (96),  (97),  in  Chap.  II. ;  Articles  (112),  (113), 
and  (114)  in  Chapter  III.  ;  the  demonstration  of  the 
rule  for  finding  the  Greatest  Common  Divisor  of  tiuo 
polynomials  in  Chap.  IV.  ;  Articles  (160)  and  (162)  in 
Chap.  VI.  ;  the  general  Discussion  of  the  Courier 
Problem  in  Chap.  X.  ;  and  the  Demonstration  of  the 
Multinomial  Theorem  in  Chap.  XVIII.  The  method 
of  solving  equations  of  the  third  and  fourth  degree 
as  set  forth  in  Articles  (335),  (337),  (354),  and  (355), 
although  tentative  in  its  character  and  not  practically 
general,  furnishes  the  means  of  resolving  many  problems 
which  have  heretofore  been  considered  difficult.  This 
method  is  considered  valuable  in  an  educational  rather 
than  in  a  scientific  point  of  view. 

An  unusual  number  of  practical  examples  have   been 
inserted  in   this  volume   on    the   principle  that  algebraic 


PREFACE.  V 

skill  can  only  be  acquired  by  extensive  practice.  Otber 
peculiarities  might  be  mentioned,  but  it  is  deemed  un- 
necessary. 

The  work  is  now  submitted  to  an  intelligent  public 
with  the  hope  that  _3fter  a  careful  examination  it  will 
meet  with  the  approbation  of  all  those  who  favor  the 
use  of  text  books  that  do  not  attempt  to  simplify  by 
the  omission  of  what  is  difficult. 


Greenmount,  near  Richmond,  Induna, 
Dec.,  1856. 


( 


CONTENTS. 

CHAPTER     I. 

DEFINITIONS. 

VXQZ 

Symbols  of  Quantity, 15 

Symbols  of  Operation, 16 

Symbols  of  Relation, IT 

Symbols  of  Aggregation, 18 

Symbols  of  Continuation, 18 

Symbols  of  Deduction, 18 

Miscellaneous  Definitions, 18 

Axioms, 21 

Exercises  in  Numeration, 22 

CHAPTER    II. 

ADDITION. 

Case  I.  and  Case  n., 23 

Casein, 25 

Case  IV., 21 

SUBTRACTION. 

Case  I., 29 

Case  IL, 30 

MULTIPLICATION. 

Propositions.     (Multiplication  of  Signs), 32,  33,  34,  and  35 

Case!, 35 

Case  XL, 36 

Case  III, 37 

Multiplication  by  Detached  Coefficients, 39 

DIVISION. 

Propositions.    (Division  of  Signs,  &c.), 40,  41,  42,  43,  44,  and  45 

Case  I., 46 

Case  XL, 4t 

Case  XXL, 48 

Division  by  Detached  Coefficients, 51 

Synthetic  Division, 52 


^"  CONTENTS. 

CHAPTEE    III. 

THEOREMS    AND    FACTORING. 

PAGB 

Theorem  I., 55 

Theorem  II., 56 

Theorem  III., 51 

Theorem  IV., 58 

Theorem  V., j,.,. ... 60 

Theorem  YL, 63 

Theorem  VII, 65 

Theorem  VIIL, 66 

Theorem  IX., 6t 

Theorem  X., 68 

Theorem  XL, TO 

Theorem  XIL, 72 

Theorem  XIIL, 14 

Problem  A. — ^To  Resolve  a  Monomial  into  Factors, 15 

Problem  B. — ^To  Resolve  a  Polynomial  into  Factors,  one  of  which  shall  be 

a  Monomial, 16 

Useful  Formulas, 18 


CHAPTER    lY. 

GREATEST    COMMON    DIVISOR. 

Theorems  L  and  n., 80 

Problem. — To  find  the  Greatest  Common  Divisor  of  Two  Polynomials, 82 

Problem. — To  find  the  Greatest  Common  Divisor  of  Two  or  more  Polynomials,     86 


CHAPTER    V. 

Problem. — ^To  find  the  Least  Common  Multiple  of  Two  or  more  Quantities, .     88 

CHAPTER    VI. 

ALGEBRAIC    FRACTIONS^ 

Theorems  I.  and  II., 90 

Problem. — To  Reduce  a  Fraction  to  its  Lowest  Terms, 91 

Problem. — ^To  Reduce  a  Fraction  to  One  Enth-e  or  Mixed  Quantity, 93 

Problem. — To  Reduce  a  Mixed  Quantity  to  a  Fractional  Form, 94 

Probleii. — To  Reduce  Fractions  to  Equivalent  Fractions  having  a  Common 

Denominator, , 96 

Problem. — To  Reduce  an  Entu-e  Quantity  to  a  Fractional  Form  having  a 

Given  Denominator, ' 98 


CONTENTS.  IX 

VAQB 

Problem. — ^To  Convert  a  Fraction  into  an  Equivalent  One  having  a  Given 

Denominator, 98 

Addition  of  Fractions, 99 

Subtraction  op  Fractions, 101 

Multiplication  of  Fractions, 103 

Division  of  Fractions, 106 

Miscellaneous  Propositions, 108 

Vanishing  Fractions, 110 


CHAPTEE    VII. 

INVOLUTION. 

Problem. — To  Raise  a  Monomial  to  the  nth.  power, 113 

Problem. — To  Raise  V^a  to  the  third  power, 114 

BINOMIAL    THEOREM. 

Problem. — ^To  Raise  a  Binomial  to  a  Given  Power, 11*7 

Problem. — To  Square  a  Polynomial, 119 

Problem. — To  Cube  a  Polynomial, 120 

Problem. — To  Raise  a  Polynomial  to  any  power, 122 


CHAPTEE    YIII. 

EVOLUTION. 

Problem. — To  Extract  a  Given  Root  of  a  Given  Monomial, 124 

Problem. — To  Extract  the  Square  Root  of —  a", 125 

Problem. — To  Extract  the  mnth  Root  of  a, 126 

Problem. — To  Extract  the  nth.  Root  of  a  Given  Quantity  to  within  a  Given 

Fraction, 128 

Problem. — To  Extract  the  Square  Root  of  a  Polynomial, 130 

Problem. — To  Extract  the  Cube  Root  of  a  Polynomial, 134 

Problem. — ^To  iBnd  the  mth  Root  of  a  Polynomial, 131 


^    CHAPTEE    IX. 


RADICALS. 
Theorems, 139 

REDUCTION    OF    SURDS, 
Problem.— To  Reduce  a  Rational  Quantity  to  the  Form  of  a  Proposed  Surd,  140 
Problem. — Reduce  a^^b  to  the  Form  of  a  Quadratic  Surd, 141 


X  CONTENTS. 

PAGB 

Problem. — To  Reduce  Two  or  more  RadicaJs  having  Different  Indices  to 

Equivalent  Ones  having  the  Same  Index, 142 

Problem. — To  Reduce  Surds  to  the  Simplest  Form, 143 

Addition  op  Radicals, 145 

Subtraction  op  Radicals, 14t 

Multiplication  op  Radicals, 148 

Division  op  Radicals, 150 

Involution  op  Radicals,  *  151 


EVOLUTION    OF  RADICALS. 

Problem. — Extract  the  mth  Root  of  a  Given  Monomial  Radical 153 

Problem. — To  Extract  the  Square  Root  of  a  Binomial  Surd,  One  of  whose 

Terms  is  Rational  and  the  other  a  Quadratic  Surd, 154 

Problem, — To  Extract  the  Cube  Root  of  a  Binomial  A-\-B, 158 

Problem. — To  Extract  any  Root  (c)  of  a  Binomial  Surd, 159 

Problem. — To  Find  such  a  Multiplier,  or  such  MultipUers,  as  will  make  any 

Binomial  Surd  Rational, 160 

Problem. — To  Reduce  a  Fraction  whose  Denominator  is  a  Surd  Quantity  to 

another  that  shall  have  a  Rational  Denominator, 163 


IMAGINARY    QUANTITIES. 

Addition  op  Imaginary  Quantitibs, 165 

Subtraction  op  Imaginary  Quantities, 166 

Multiplication  op  Imaginary  Quantities, 167 

Division  op  Imaginary  Quantities, ItO 

Involution  op  Imaginary  Quantities, 176 

Evolution  op  Imaginary  Quantities, 172 

Modulus, 174 

Theorems, 175 

Miscellaneous  Examples  in  Radicals, 176 


CHAPTER    X. 

EQUATIONS. 
Theorems, 179,  180,  181,  and  182 

SIMPLE    EQUATIONS. 

Simple  Equations  Containing  One  Unknown  Quantity, 

Examples, 189 

Questions  Involving  Simple  Equations  Containing  One  Unknown  Quantity, .   197 


CONTENTS.  XI 

PAGB 

Simple  Equations  op  the  Fiest  Degeee  Containing  Two  Unknown 
Quantities, 209 

ELIMINATION. 

Elimination  by  Substitution, 211 

Elimination  by  Comparison, , 214 

Elimination  by  Addition  and  Subtraction, 211 

Miscellaneous  Examples  in  Elimination, 220 

Simultaneous  Equations  of  the  First  Degree  Containing  Three  or  Mpre  Un- 
known Quantities, 224 

Questions  involving  Simultaneous  Equations  of  the  First  Degree, 235 

Interpretation  of  Negative  Eesults, 242 

General  Discussion  of  Certain  Results  Obtained  in  the  Solution  of  Simple 

Equations, 245 

COURIER    PROBLEM. 

Examples, 255 

Simple  Inequalities, 266 


CHAPTEE    XI. 

QUADRATIC    EQUATIONS. 

Pure  Quadratics, 261 

Simultaneous  Equations, 264 

Questions  Producing  Pure  Quadratics, 266 

Affected  Quadratics, 268 

Examples, 270 

Examples, 275 

Examples, 278 

Examples, 279 

Examples, 281 

Examples, 282 

Examples, 284 

Examples, ^. 285 

Examples, 286 

Examples, 287 

Examples, 288 

Examples, 290 

Examples, , 291 

Examples, 292 

Miscellaneous  Examples, 293 

Affected  Quadratics  Involving  Two  Unknown  Quantities, ....  297 

Questions  Producing  Affected  Quadratics, 299 


XU  CONTENTS. 

CHAPTEE    XII. 

CUBIC    EQUATIONS. 

PAGB 

Pure  Cubic  Equations  and  Examples, 30*7 

Incomplete  Cubic  Equations  and  Examples, 308 

Examples, 310 

Examples, 311 

Examples, 3.12 

Henkle's  Method, 313 

Examples, 316 

Questions, 319 


CHAPTEE    XIII. 

BIQUADRATIC    EQUATIONS. 

Pure  Biquadratics, : 322 

Affected  and  Incomplete  Biquadratics, 324 

Examples, 325 

Biquadratics  Containing  Two  Unknown  Quantities, 339 

Examples, 340 

Miscellaneous  Questions, 346 


CHAPTEE    XIY. 

HIGHER     EQUATIONS. 

Problems, 348 

Examples, 350 

Examples, 352 

Simultaneous  Equations, ■ 355 


CHAPTEE    XY. 

ARITHMETICAL    PROGRESSION. 

Problems, 358 

Propositions, 359 

Questions, ?*. 359 

Questions, 362 


CHAPTEE    XVI. 

GEOMETRICAL    PROGRESSION; 

Problems, 365 

Propositions, 367 


■1 


CONTENTS.  xiii 

PAGB 

Examples, 36t 

Questions, 371 


CHAPTEK    XYII. 

PROPORTION. 

Introductory  Definitions  and  Explanations, SYS 

Propositions  and  Demonstrations, 376 

Harmoxical  Proportion, 382 

ilarmonical  progression, 383 

Examples, 386 

Problems  and  Questions  in  Proportion, 388 


CHAPTER    XYIII. 

SERIES. 

Expansion  of  Series, 392 

The  Method  of  Undetermined  Coefficients, 394 

Decomposition  of  Rational  Fractions, 399 

Examples, 400 

Method  of  Involution, i 402 

Method  of  Evolution, 402 

Binomial  Theorem, 402 

Examples, 407 

Multinomial  Theorem, 408 

Reversion  of  Series, 412 

Examples, 414 

Summation  of  Series, 415 

Examples, 423 

The  Differential  Method, 426 

Examples, 428 

Apphcation  of  the  Differential  Method, 430 

SPECIAL    SERIES. 

Questions, 434 

Promiscuous  Questions 439 


ALGEBRA. 


CHAPTER  I. 
DEFINITIONS. 

(1.)  Algebra  is  a  general  metliod  of  investigating  the  relations  of 
quantities. 

(2.)   Quantity  is  that  which  admits  of  increase,  or  of  diminution. 

(3.)  Algebraic  notation  is  the  method  of  representing  by  symbols 
algebraic  quantities,  their  relations,  and  the  operations  to  be  performed 
upon  them. 

(4.)  Algebraic  symbols  are  characters  used  in  algebraic  expressions. 

(5.)  Algebraic  symbols  are  of  six  kinds ;  namely, 
Symbols  of  Quantity, 
"         "   Operation, 
"         "  Relation, 
"         "  Aggregation, 
"         "  Continuation, 
and    "        "  Deduction, 

SYMBOLS  OF  QUANTITY. 

(6.)  The  symbols  of  quantity,  generally  used,  are  the  Arabic 
numerals  and  alphabetic  characters. 

(7»)  The  Arabic  numerals,  1,  2,  3,  4,  5,  <fec.,  are  used  to  represent 
known  quantities. 

(8«)  The  first  letters  of  the  alphabet,  a,  b,  c,  &c.,  are  generally 
used  to  represent  known  quantities,  or  quantities  that  may  be  assumed 
to  be  of  any  value  whatever. 

(9.)  The  final  letters  of  the  alphabet,  x,  y,  z,  u,  &c.,  are  generally 
used  to  represent  quantities  which  depend   for  their  values  upon 


16  DEFINITIONS. 

known  quantities,  or  upon  quantities  which  may  be  assumed  to  be 
known. 

( 1 0.)  The  symbol  0  is  called  zero^  and  denotes  the  absence  of  quan- 
tity, or  that  which  is  less  than  any  assignable  quantity. 

(11.)  The  symbol  oo  is  called  infinity^  and  denotes  that  which  is 
greater  than  any  assignable  quantity. 

(12.)  The  notations,  a',  a",  a'",  a"",  &c.,  and  «,,  a,,  «3,  «4, 

a„,  are  often  used  to  denote  different  quantities  which  occupy  similar 
positions  in  different  operations,  a!  is  read  a  'prime ;  a",  a  second ; 
a"',  a  third  ;  a"",  a  fourth  ;  &c.,  and  ctj,  is  read  a  sub  one  ;  a^^a  sub 
two  ;  ttg,  a  sub  three  ;  a^^a  sub  four a„,  a  sub  n. 

(13.)  The  symbols  ',  '',  "',  "",  &c.,  are  called  accents.  The  symbols 
I,  2,  3,  4, nj  are  called  subscripts. 

SYMBOLS  OF  OPERATION. 
(14.)  The  symbols  of  operation^  are  +,  — ,  '^,  x,  ',  -f-,  :,  — ,  ), 

i_,  ^  ^  \  &c.,  2 '  3 .  h  &c.,  |/,  V,  V,  &c. 

(15.)  The  symbol  +  is  called  plus,  and  is  the  sign  of  addition. 
Thus,  a  +  b  indicates  the  addition  of  a  and  b. 

(16.)  The  symbol  —  is  called  minus,  and  is  the  sign  of  subtrac- 
tion.    Thus,  a— b  indicates  the  subtraction  of  b  from  a. 

(17.)  The  symbol  ^,  when  placed  between  two  quantities,  denotes 
that  the  less  is  to  be  subtracted  from  the  greater.  Thus,  6~5,  or 
5 '^6,  denotes  that  5  is  to  be  subtracted  from  6  ;  and  a'^b  denotes  that 
6  is  to  be  subtracted  from  a,  or  a  from  6,  according  as  a  is  greater  or 
less  than  b. 

(18.)  The  s)rmbols  x  and  *  arie  signs  of  multiplication.  Thus, 
7x5,  or  Y-S,  indicates  that  7  is  to  be  multiplied  by  5,  or  5  by  7 ; 
a  X  6,  ov  a-b,  indicates  that  a  is  to  be  multipHed  by  6,  or  b  by  a. 

(19.)  In  representing  the  multiplication  of  literal  quantities,  the 
sign  of  multiplication  is  generally  omitted.  Thus,  instead  of  a  x  6,  or 
a-6,  we  write  ab ;  and  instead  of  2  x  a,  or  2-a,  we  write  2a. 

(20.)  The  symbols  ^,  :,  — ,  ),  and   | ,  are  signs  of  division. 

Thus,  6-^3,  6:3,  f,  3)6,  or  6|3  ,  indicates  the  division  of  6  by  3 ; 

also,  a -4- 6,  a :  6,  7^,  6) a,  or  a|6  ,  indicates  the  division  of  a  by  b. 


BEFnanoiTs.  17 

(21*)  The  symbols  \  ',  ',  *,  &c.,  are  called  ex2yonents,  and  are 
signs  of  involution.  Thus,  a*,  which  is  an  abbreviation  of  aaaaa, 
denotes  that  a  is  to  be  raised  to  the  fifth  power,  that  is,  a  is  to  be  taken 
five  times  as  a  factor.     The  exponent  *  is  not  usually  written. 

(22.)  The  symbols  2»  3">  4j  ^q.,  are  called  fractional  exponents, 

and  are  the  signs  of  evolution.  Thus,  a^,  a^",  a*,  &c.,  respectively 
indicate  that  the  square  root,  the  cube  root,  the  fourth  root,  &c.,  of  a 
are  to  be  extracted. 

(23.)  In  fractional  exponents  the  numerator  denotes  a  power,  and 

3. 

the  denominator  a  root.  Thus,  x^  denotes  the  fifth  root  of  the  third 
power  of  X,  or  the  third  power  of  the  fifth  root  of  x. 

(24.)  The  symbol  V  is  called  the  radical,  and  is  the  sign  of  evolu- 
tion. Thus,  V^,  V«,  Vflt,  &c.,  respectively  indicate  that  the  square 
root,  the  cube  root,  the  fourth  root,  &c.,  of  a  are  to  be  extracted. 
The  small  figure  in  the  angle  of  the  radical  is  the  index  of  the  root. 
When  no  index  is  written, "  is  understood.  Thus,  Va  is  the  same  as 
Va. 

SYMBOLS  OF  EELATION. 

(25.)  The  symbols  of  relation  are  :,  =,  : :,  >  <,  -f-  .  .  ., 
and  -H-     :     :     :     . 

(26.)  The  symbol  :  denotes  ratio.  Thus,  a  :  b  denotes  the  ratio 
of  a  to  b. 

(27.)  The  symbols  =  and  :  :  are  signs  of  equality.  Thus,  a=b 
denotes  that  a  equals  b ;  and  a:b=c:d,  ot  a:b  ::c:d  denotes  that 
the  ratio  of  a  to  6  equals  the  ratio  of  c  to  d. 

The  symbol  :  :  is  not  used  except  to  denote  the  equality  of  ratios. 
Thus,  we  never  write  a::b  hr  a=b. 

(28.)  The  symbols  >  and  <  are  signs  of  inequality.  Thus, 
a>6  denotes  that  a  is  greater  than  b  ;  and  a<6,  that  a  is  less 
than  b. 

(29.)  The  symbol  -i-  ...  is  the  sign  of  an  arithmetical  series. 
Thus,  -4-  a .  6 .  c .  c? denotes  the  equality  of  the  difference  between  a  and 
b,  b  and  c,  and  c  and  d, 

(30.)  The  symbol  -^f  :  :  :  is  the  sign  of  a  geometrical  series. 
Thus,  -~a:b:c:d  denotes  the  equality  of  the  ratios  of  a  to  6,  6  to  c, 
and  c  to  d. 

2 


18  DEFINITIONS. 

SYMBOLS  OF  AGGEEGATION. 
(31»)  The  symbols  of  aggregation  are  ,  | ,  {  ),  [  ],  and  \    j-. 

(32#)  The  symbol  is  called  a  vinculum,  and  denotes  that  the 

quantities  over  which  it  is  placed  are  to  be  considered  as  one  quantity. 
Thus,  Va  +  h-\-c  denotes  that  the  square  root  of  the  sum  of  a,  6,  and  c 
is  to  be  extracted. 

(33»)  The  symbol  |  is  called  a  6ar,  and  denotes  that  the  quantities, 
in  the  column  immediately  preceding  it,  are  to  be  considered  as  one 

-\-ax 
quantity.     Thus,    +  h     denotes  that  the  sum  of  a,  6,  and  c  is  to  be 

+  c 
multiplied  by  x. 

(34.)  The  parenthesis  (  ),  brackets  [  ],  and  braces  -j  j-,  denote 
that  the  quantities  contained  within  them  are  to  be  considered  as  one 
quantity.  Thus,  (6  +  c)ar  denotes  that  the  sum  of  b  and  c  is  to  be 
multiplied  by  x;  la-\-{b  +  c)x^g  denotes  that  the  sum  of  a  and 
(6  4- c)x  is  to  be  multiplied  by  y;  and  •j2  +  [a  +  (6  +  c)ic]y|'tt  denotes 
that  the  sum  of  z  and  [a  +  (6  +  cjx^y  is  to  be  multiplied  by  u. 

SYMBOLS  OF   CONTINUATION. 

(35.)  The  symbols  of  continuation  are  ....  and ,  and  are 

equivalent  to  c&c,  and  so  on,  or  continued  according  to  tike  same  law. 
Thus,  a,  a'*,  a^,  a*,  a^,  a",  a',  a®,  &c.,  and  a^,  a^,  a^,  a^,  a^,  a„,  a„.  a 
&c.,  may  be  written  a,  a",  a^  .  . .  .  and  a^,a^,a 


g,     .     .     .     . 


SYMBOLS  OF  DEDUCTION. 
(36»)  The  symbols  of  deduction  are  .• .  and  ••• 

(37.)  The   symbol    .*.   signifies    therefore,   whence,  ccmsequentlpy 
hence,  from  which  we  infer,  <fec. 

(38.)  The  symbol  *  .•  signifies  simie,  or  because. 

(39»)  A  monomial,  or  term^  is  an  algebraic  expression  which  is  not 
connected  to  any  other  by  the  sign  of  addition  or  subtraction ;  as,  a, 

2a,  ab,  aHc,  ^,  (a-\-b),  {x-\-y)x,  {x-\-a)  (y  +  6),  &c. 

(40.)  A  binomial  is  an  algebraic  expression  which  is  composed  of 
two  terms;  as,  a +  6,  2ar— 3y,  {x-{-y^-\-z,  (a;  +  a) -f- (y-|-6),  <fec. 


DEFINITIONS.  19 

(41«)  A  trinomial  is  an  algebraic  expression  which  is  composed 
of  three  terms ;  as,  a  +  6  +  c,  a?— y  +  32,  {x-\-y)-\-a-\-{c-\-d)^  (^~y) 
—  (a— 6)4-(c— c?),  &c. 

(42.)  A.  polynomial,  or  multinomial^  is  an  algebraic  expression 
which  is  composed  of  several  terms;  as  a +  5,  a—h-\-c,  x  +  y—z+n, 
a  +  b—c  +  d—e,  &c. 

(43.)  A  residual  is  an  algebraic  expression  which  denotes  the 
difference  of  two  terms;  as,  a—b,  (a  +  6)— (c  +  c?),  &c. 

(44.)  A  simple  term  is  one  which  contains  but  one  sign  of  addi- 
tion or  subtraction  expressed  or  understood;  as,  —Bar,  4a6,  &c. 
But  (a  +  b)  is  not  a  simple  term,  because  it  is  equivalent  to  (  +  a4-6) 
which  contains  two  plus  signs. 

(45.)  A  compound  term  is  one  which  is  composed  of  two  or 
more  simple  terms  affected  by  a  sign  of  aggregation;  as,  (a— 6), 
V2x—3y  +  4:Z,  \(x-hy)z+  (a  +  b)  [c+  (m—ny]-\-q\,  &c. 

(46.)  A  positive  term  is  one  that  is  affected  by  the  sign  of  addi- 
tion ;  as,  4- 6a;,  4- («  +  &),  +(^— y),  or  6a?,  (a +  5),  (x—y),  &c. 

(47.)  A  negative  term  is  one  that  is  affected  by  the  sign  of  sub- 
traction; as,  —6a,  —  (a?— y),  —{Bx—4y-{-1z)j  &c. 

(48.)  Like  J  or  similar  terms,  are  those  that  contain  the  same 
letters  affected  by  the  same  exponents ;  numerals  denoting  abstract 
numbers ;  and  numerals  denoting  concrete  numbers  of,  the  same  de- 

nomination.  Thus  6a'a;'  and  69aV,  and  4:Xy^  and  —Ixy^,— 
6(a'  +  6')  and  42(a''  +  6'),  61  and  34,  $12  and  —$10  are  respectively 
similar. 

(49.)  Unlike,  or  dissimilar  terms,  are  those  that  are  not  similar. 
Thus,  a  and  x,  a^x  and  aa;",— 6(a;  +  y)  and— '7(a;+2),  are  respectively 
dissimilar. 

(50.)  Homogeneous  terms  are  those  in  which  the  sum  of  the  ex- 
ponents of  the  literal  factors  in  each  are  equal.  Thus,  2x^y  and  xy^, 
4:a^b^  and  5ab\  are  respectively  homogeneous. 

(51.)  The  coefficient  of  a  term  is  that  factor  which  is  considered 
as  denoting  the  number  of  times  the  rest  of  the  term  is  taken.  Thus, 
in  6a,  6  is  the  coeflScient  of  a  ;  in  ax,  a  may  be  considered  the  coeffi- 
cient of  a:,  or  1  may  be  considered  the  coeflScient  of  ax  ;  also,  in  6aaJ, 
6a  is  the  coeflScient  of  x,  and  6  the  coeflScient  of  ax. 


20  DEFINITIONS. 

(52.)  ^  numerical  quantity/  is  one  whicli  is  expressed  by 
numerals  ;  as,  6,  43,  &c. 

(53.)  A  literal  quantity  is  one  which  is  expressed  entirely  or  in 
part  by  letters  ;  as,  ax^  Sab^  &c. 

(54.)  A  rational  quantity  is  one  which  can  be  expressed  without 

the  aid  of  a  radical  sign  or  fractional  exponent ;  as,  a,  \/a^=za,  h^  = 
6',v/36  =  6,  &c. 

(55.)  An  irrational  quantity  is  one  which  can  not  be  expressed 

without  the  aid  of  a  radical  sign  or  fractional  exponent ;  as,  v'o,  6^, 
/2,  &c. 

(56.)  The  power  of  a  quantity  is  the  product  resulting  by  taking 
the  quantity  a  certain  number  of  times  as  a  factor.  Thus,  aaaa  or 
a*  denotes  the  4th  power  of  a. 

(57.)  The  root  of  a  quantity  is  a  quantity  which  taken  a  certain 
number  of  times  as  a  factor,  produces  the  given  quantity. 

(58.)  The  reciprocal  of  a  quantity  is  unity  divided  by  that  quantity. 

Thus,  -  is  the  reciprocal  of  a,  and  -,  or  -  is  the  reciprocal  of  ■^. 
a  ^  a       a  ^  b 


(59.)  An  algebraic  formula  is  the  general  expression  of  a  mathe- 
matical truth. 

(60.)  A  proposition  is  something  proposed  to  be  done,  or  to  be 

demonstrated. 

(61.)  A  problem  is  something  proposed  to  be  done. 
(62.)  A  theorem  is  something  proposed  to  be  demonstrated. 

(63.)  A  lemma  is  something  to  be  demonstrated  in  order  to  ren- 
der what  follows  more  easy. 

(64.)  A  corollary  or  consectary  is  an  obvious  consequence  deduced 
from  some  preceding  truth  or  demonstration. 

(65.)  A  scholium  is  a  remark  appended  to  the  demonstration  of 
a  theorem  or  to  the  solution  of  a  problem. 

(66.)  An  hypothesis  is  a  supposition  made  in  the  enunciation  of 
a  propositiqn,  or  in  the  course  of  a  demonstration. 


DEFINITIONS.  21 

(67.)  A  direct  demonstration  is  one  in  which  the  reasoning  em- 
ployed is  regular  deduction. 

(68.)  An  indirect  demonstration  is  one  by  which  a  thing  is 
proved  to  be  true  by  showing  that  the  supposition  that  it  is  not  true 
leads  to  an  absurdity.  This  kind  of  demonstration  is  frequently  called 
a  reductio  ad  ahsurdum. 

(69.)  An  axiom  is  a  self-e\ddent  truth. 

AXIOMS. 
1.  The  whole  is  equal  to  the  sum  of  all  its  parts. 

2t  If  equal  quantities  be  added  to  equal  quantities,  the  sums  will 
be  equal. 

3.  If  equal  quantities  be  subtracted  from  equal  quantities,  the 
remainders  will  be  equal. 

4 1  If  equal  quantities  be  multiplied  by  equal  quantities,  the  pro- 
ducts will  be  equal. 

5.  If  equal  quantities  be  divided  by  equal  quantities,  the  quotients 
will  be  equal. 

6«  If  unequal  quantities  be  added  to  equal  quantities,  the  sums  will 
be  unequal. 

It  If  unequal  quantities  be  subtracted  from  equal  quantities,  the 
remainders  will  be  unequal. 

8.  If  unequal  quantities  be  multiplied  by  equal  quantities,  the 
products  will  be  unequal. 

9.  If  unequal  quantities  be  divided  by  equal  quantities,  the  quo- 
tients will  be  unequal. 

lOt  Quantities    which  are  an  equal   number  of  times   the    same 
quantity,  are  equal  to  each  other. 

11,  Equal  powers  of  equal  quantities  are  equal. 

12.  Equal  roots  of  equal  quantities  are  equal. 

(70.)  EXERCISES  IN  NOTATION. 

1.  Write  a  added  to  6. 

2.  Write  a  subtracted  from  h. 

3*  Write  the  difference  between  a  and  ft.  Ans,  a^^h. 


22  AXIOMS. 

I.  Write  in  three  ways  a  multiplied  by  h, 
5«  Write  in  two  ways  6  multiplied  by  8. 
0*  Write  in  five  ways  a  divided  by  6. 

7.  Write  in  three  ways  the  product  of  a  added  to  h  by  the  differ- 
ence between  h  and  a  when  a  is  greater  than  6. 

8.  Write  in  two  ways  the  square  root  of  a,  added  to  h. 

.  9.  Write  in  two  ways  the  cube  root  of  a,  added  to  the  square  of 
the  sum  of  a,  6,  and  c. 

lOi  Write  in  two  ways  the  square  root  of  the  sum  of  a  and  6. 

,11.  Write  in  two  ways  the  cube  root  of  the  sum  of  a  and  the 
square  of  the  sum  of  a,  6,  and  c. 

12.  Write  the  product  of  the  sum  of  a  and  h  by  the  sum  of  x  and 
y,  divided  by  the  cube  root  of  a,  diminished  by  d, 

(71.)  EXERCISES  m  NUMERATION, 

1.  Read  a +  6.  ^ 

2.  Read  a— 5. 
3i  Read  a'^c, 

I.  Read  6  x  7  or  6*7. 
5*  Read  a  x  5,  or  a'h, 
6*  Read  ah. 

7.  Read  a-^b. 

8.  Read  a:6.  Ans.  a  divided  by  6,  or  the  ratio  of  a  to  6. 
0.  Read  a*.  Ans.  a  squared,  a  square,  or  aV  square. 

10.  Readl^a. 

11.  ReadVa+y. 


\/a- 


12.  Readi/a+l^ft  +  c. 


13.  Reader  «+^6 4- Vc. 


14.  Read  3  \/ a^ -^rVa^ +Va. 

15.  Read  (x'-y'')  |  a  +  6[a  +  3(6-i>)]  +  4y»  I  ^ 


V'    .       o^^    •       ^'^ 


CHAPTER   II. 

\^        ^       ADDITION. 

(72.)  Actdition  is  finding  the  simplest  expression  for  the  sum  of 
several  algebraic  quantities. 

CASE    I. 
(73.)  When  quantities  are  entirely  dissimilar. 

RULE. 
Connect  them  together  by  their  proper  signs. 


PROBLEM. 

Find  the  sum  of  a,  —6,  +3c,  and  —5d, 

SOLUTION. 

Connecting  these  expressions  together,  we  have,  a— 6  +  3c~6(?. 

EXAMPLES. 

!•  Find  the  sum  of  Qa  and  —'76.  Ans.  6a— lb. 

2.  Find  the  sum  of  4a,  —36,  and  -\-5x.  Ans.  4a—3b-{-5x. 

3*  Find  the  sum  of  aa;,  +bm,  — cy,  and  -{-nw. 

Ans.  ax  +  bm—cg+nw. 

CASE    II. 
(7  4.)  AVTien  the  quantities  are  similar  and  have  the  same  sign. 

RULE 

1.  When  the  quantities  are  numerical,  add  as  in  arithmetic,  and 
prefix  the  sign  +,  or  —,  as  the  case  mag  be. 

RULE 

2.  When  the  quantities  are  literal,  add  the  coefficients,  affix  the  literal 
part,  and  prefix  +,  or  —,  as  the  case  mag  be. 

PROBLEM 

1.  Find  the  sum  of  —4,  —1,  —8,  and  —3. 


24  ADDITION. 

SOLUTION. 

Operation. 

—  4  Adding  these  numerals  together  and  prefixing  the 

—  1  sign  — ,  we  have  —22. 

—  8 

—  3 


-22 

PROBLEM 

2.  Find  the  sum  of  —Zahx,  —lahxj  —GabXj  and  —9ahx, 


Operation, 

—  Sabx 

—  labx 

—  6abx 

—  dabx 

SOLUTION. 

Adding  the  numerical  coefficie 
sign  — ,  and  annexing  the  comn 
we  have  —25abx. 

mts,  prefixing  th< 
ion  literal  factor 

— 

25abx 

EXAMPLES, 

1. 

2. 

3. 

4. 

5. 

4a 
a 

4a6 
lab 

—  abc 

—  Sabc 

5{aJrh) 
Ha  +  b) 

2|/a+a;' 
dVa  +  x^ 

6a 

bob 

—  4abc 

S{a  +  b) 

5Va-\-x^ 

a 

2ab 

—  12abc 

U{a  +  b) 

1Va  +  x' 

2a 

ab 

-  2abe 

(a  +  b) 

Wa+x^ 

8a 
8a 

*25a 

] 

dab 
llab 

33ab 

-Uabc 

12{a  +  b) 

lOKa  +  ic' 

—BQabc 

42(a  +  6) 

6. 

7. 

8. 

-4(a' 

-t^)i 

Bix^+y)i 

4(.r'-y«)* 

-  V(a' 
-11  (a» 

-b')i 

Wx^+y 
Wx'-^y 

QVx'-y' 

Ql^x'-y'' 

-  8(a' 

-6^)* 

l^ix'^yY 

Vx^-f 

-  3(a' 

-b')^ 

lQ{x'  +  yY 

^{x^-f)^ 

*  When  no  sign  is  written  +  is  understood. 


9. 

ADDn 
10. 

3a;V— 4y' 

6a;'y— 3y' 
8a:'y— 4y' 

11. 

3a' +     p 

{a^^hy-\QVrrv'-n^ 

Ya'+   3p 

^Va'-h-"    -  AVm'-n^) 

9a' +    6i? 
14a'  +  16i? 

^(a'-^Y-  5(m'-ii')2 

25 


12. 

3a'—  7  +  5a6c 
4a'—  6  +  7a5c  +  4a;'y 
5a'—   3  +  8a6c  +  5a;''y  +  4m7i 
6a'—  4:  +  Qahc  +  Qx^y+   mn+   cd 
7a' — 1 2  4-  3a6c  +  4a;'y  +  Smn  +  4cd 

CASE     III. 
(75.)  When  the  quantities  are  similar,  and  all  have  not  the  same 


EULE. 
Mnd  the  sum  of  the  similar  positive  terms,  also  the  sum  of  the 
similar  negative  terms,  and  then,  disregarding  the  signs,  ascertain 
their  difference,  and  prefix   +,  or   —,  according  as  the  sum  of  the 
positive  or  negative  terms  is  greater, 

PROBLEM.       . 

Find  the  sum  of  6a,  —*la,  —3a,  +4a,  +2a.  — ? 

SOLUTION. 

Operation. 

+  Qa  The  sum  of  the  positive  terms  is   +10a,  and  the 

—  Ya  sum  of  the  negative  terms  is  —12a.     Disregarding 

—  3a  the  sio^ns,  we  have  2a  for  the  difference  between  12a 
+  4a  and  10a,  to  which  prefix  — ,  because  12a  is  greater 
—2a  than  10a  and  has  a  minus  sign. 

—  2a  , 

"We  may  also  add  the  terms  successively.  Thus,  —  2a  +  4a  is 
+  2a;  which,  added  to  —3a,  is  —a;  which,  added  to  —7a,  is  —  8a^' 
which,  added  to  -\-6a,  is  —2a. 


26  ADDITION, 

EXAMPLES. 

!•  What  is  the  sum  of  —  4a;y,  -^^xy^  +6a;y,  and  —bxy? 

Ana,  Axy, 

2t  What  is  the  sum  of  —  6a6,  —  taft,   +3a6,  and  -f  4a6.^ 

-4w5.  —  6a&. 

3*  What  is  the   sum  of  3a+xy,    —^a—4txy,    +1a—5xyj   and 
Qa  +  xy?  Ans.  I2a—1xy, 

4.  What  is  the  sum  of  4,xy—ab^  —Bxy  +  4abj   —Axy—bab^  and 
+  5xy  +  4a6  ?  ^ws.  2a;y  +  2ah. 

5*  What  is  the   sum  of  a  +  6  +  c+c?  +  e— /,   a  +  6  +  c  +  (f— e4-^ 
a  +  6  +  c— c?  +  e+/,  a-\-h—c-\-d  +  e+f,  a— 6  +  c  +  e4-/,  and   — a  +  J 

6«  What  is  the  sum  of  7a— 5c +  36  and  2a— 3c— lb  ? 

Ans.  9a— 8c— 46. 

7t  What  is  the  sum  of  6a  +  46— 3c— Yc^  +  S  and  3a— 126  +  7c— 
lOrf— 4?  Ans.  8a— 86  +  4c— l7rf  +  4. 

8.  What  is  the  sum  of  —  Y/+3a,  4/— 2a,  3/— 3a,  and  +2a.^ 

Ans.  0. 

9t  What  is  the  sum  of  12A— 3c— 7/+3^  and  —  3^4-8c— 2/— Q^r 
-\-bx?  Ans.  9A  +  6C— 9/— 6^  +  6a;. 

10.  What  is  the  sum  of  16a— 66  + 15c— 9c?,  3a  +  186— 5c— Ydf+ 
3c,  — 7a— 26— 3£?  +  5e— 9A,  and  11a— 36  +  2c  +  8c?  +  7A.^ 

Ans.  23a  +  86  +  12c— llc?+8c— 2A. 

U,  What  is  the  sum  of  8a  +  6,  2a— 6  +  c,  — 3a  +  56  +  2c?,  —66— 3c 
+  3c?,  and  —  5a  +  7c— 2c?.^  Ans.2a—b-\-5c-\-3d. 

12.  What  is  the  sum  of  — a  +  36— c— 115<?  +  6c— 5/,  3a— 26— 3c 
~-d  +  2le,  56— 8c  +  3c— 7/,  7a— 66  +  l7c  +  9c?— 5e  +  ll/,  and  —3a 
— 5c— 2c?  +  6e— 9/+^/  Ans.  6a— 109<?  + 37c— 10/+^. 


13.  What  is  the  sum  oi  Vx''  +  y''—m''-\-ri'—2mn,--Vx'+y^  +  3m^ 

—Bn"  +  5ww,— 5l/?  +  y"— 4m'  +  5?^'- 7m7i,  2(a;'  +  y'')  a  +  12m'— 2  Jn' 

+mw,  and  8(ar'+y')2 -8m^-iw''-6m?i  2        

Ans.  5f^a;'+y'  +  2m'— 9ww. 


14.  What  is  the  sum  of  5ax^ —Vx-^y+{a-'b),  —1aVx  +  2(x-\-y)l 


I 


ADDITION.  27 


— 3(a— 6),  12aVx—3Vx  +  y  +  12(a-'b),—Ba\^x—'iVx-\-y—{a—b), 
and— aa?i -F- (a;+y)i— 3(a— 6)  ? 

Ans.  Qaxi—5Vx+y  +  6{a—b). 


15.  What  is  the  sum  of  2Vxy+xz  +  j/z  +  Vax  +  bi/,  5Vxi/+xz-\-yz 
—  S{ax  +  by)ij  12  (xy  +  xz +  yz)^^ +  5  {ax  ■\-by)l,  —3Vxy-^xz+yz—2 
Vox  +  by^  and  (xy  -\-xz  +  yz)^  +  (ax  -{-byY'i 


Ans,  1 7  Vax  ->rXZ+yz-\-  2\/ax  +  by. 


16.  What  is  the  sum  of  ^{a-^b)Vx''-y''-2{a-b)Vx''  +  y\  -3(a 
^, 6) f^d^''  +  {a-b)Vx-'Tf,  -{a-Vb){x'-f)UMP'-h){x''  +  f)\, 
%{a  +  b){x''-y'')\-{a-b){x'  +  y'')\,lO{a  +  b)Vx'-y^-5(a-b)(x^  + 
y')i,  and  -2(a^b){x^-y^)\  +  4.{a-b)Vx''-\-y'"i 


Ans,  \4:(a-\-b)Vx''-y\ 

17.  What  is  the  sum  of  10f^2  +  5V8-7V5  +  2Va,  6i/2  +  V8  + 
4V5-3l/a,  and-3i/2-9l/8-3V5  +  V«+i^a^; 

^w*.  12f2  — 3V8— eVS+f'aS 

18.  What  is  the  sum 'of  6a*6  +  3a-'6'c— Ya6,    —  6a*6  +  2a-'6V^- 
l7a6,  and  9a*6— 8a-'6'c— lOaS;  ^n».  Sa'b—3a-''b\, 

19.  What  is   the   sum   of    —3(ax-{■by->^cz)^-^^x'^\■y''^\^a—b,   2 

V^^+lyT^+(a:'  +  y')^-3  {a-b),     \/ax+by  +CZ-  fVTy'  +  2 

(a-^>),3Vaa;+  6y  +cz  +  {x'-\-y'')\-{-a-b,6Vax+  by  +cz  +  {x'+y'')i 

-2{a-b),  and  {ax  +  by-\-cz)\-Vx^  +  y''-3{a-b)'i 

Ans,  Q{ax-\-by-\-cz)\—i:{a—b), 

20.  What  is  the  sum  of  Gasftf— 9cf<?4-10ai6f,  — Gaifti— a^6§  +  6 
cK  2cf(?-3ai54--3a56f,  and  -2albl^cld-^a\b^'\  Ans,  0. 

CASE    IV. 

(76.)  When  the  quantities  are  partially  similar,  and  have  the 
same  or  different  signs. 

RULE. 

Find  the  expression  for  the  sum  of  the  coefficients  of  the  terms,  and 
annex  a  part,  or  all  of  what  is  common  to  each  term,  according  as  the 
addition  is  to  be  partial  or  complete. 

Remark. — We  may  take  as  many  factors  of  any  term  for  the  coeflBciont  as  we 
choose. 


28  ADDITION. 

P  B  O  B  L  E  M 

1.  Find  an  expression  for  the  partial  addition  of  6max,  Snax, 
12paXj  and  9vax, 

SOLUTION. 

Considering  x  as  the  common  factor,  the  coefficients  are  6ma,  Sna, 
12pa,  and  9va,  whose  sum  may  be  represented  by  {6ma  +  Sna-\-12pa 
+  9va),  to  which  annexing  x,  we  have  {6max  +  3na -{■  I2pa -\- 9va)x 
for  an  expression  of  the  partial  addition  of  Qmax  +  Snax  +  12pax  + 
9vax. 

Considering  ax  as  the  common  factor,  the  coefficients  are  6wi,  3w, 
12/),  and  9v,  whose  sum  maybe  represented  by  (Qm -\- 3n  +  12p •{- 
9v),  to  which  annexing  ax,  we  have  {Qm  +  3n -\- 12p  +  9v)ax  for 
another  expression  of  the  partial  addition  of  Qmax  +  Snax-^12pax 
+  9vax, 

PROBLEM 

2.  Find  an  expression  for  the  complete  addition  of  Qmax^  Snax 
12paXj  and  9vax, 

SOLUTION. 

We  observe  that  6max  =  2m'Sax, 

Snax  =    n'SaXj 
\2pax  =    4p*3aar, 
and  9vax  =    3v'3ax, 

From  this  we  see  that  3ax  is  the  entire  common  factor  to  which 
prefixing  the  expression  for  the  sum  of  the  coefficients  2m,  n,  4p,  and 
3?;,  gives  (2m  +  n-\-Ap  +  3v)Sax,  or  3(2m  +  n  +  4:p-\-  3v)  ax  for  the  ex- 
pression of  the  complete  additon  of  Qmax,  3nax,  \2pax,  and  9vax, 

EXAMPLES. 

1,  Find  the  sum  of  aar,  6a:,  and  car.  Ans,  {a  +  h-\-c)x,    . 

2«  Find  the  sum  of  ax-\-by  +  cz,  hx-\-cy-{-  az,  and  cx  +  ai/-{-  bz. 

Ans.  (a-\-b-\-c)(x-{-y  +  z). 

3*  Find  the  sum  of  Qx^y  +  1x^z  +  9x^ym, 

Ans,  [3(2  +  3m)y  +  Y2;]a;'. 

4.  Find  the  sum  of  (a—b)Vx  +  (m—n)Vi/+V2,  (a-\-c)ri—{m—n) 

y^-h2^2,{b—cyx  +  3(m—n)Vy—3V2,and(c—a)Vx—5(m—ny'y— 
eV2,  Ans.  {a-{-c)^x—2{m-n)Vv-eV2. 


ADDITION.  29 

5*  Find  the  sum  of  (m  +  n)y^—(a—b)x^-}-axi/j  {n—p)y^—(2a-{- 
h)x^—bx7/,  (^— 2%'— (c— 3a)  a;'  +  cary,.  and  (q—m)y^—{c  +  2d)x^ 
—dxy.  Atis.  qy^—2{c  +  d)x^  +  (a—b  +  c—d)xy, 

6*  Find  the  sum  of  ax'^  +  by  +  c  and  dx^-\-hy  +  k. 

Ans.  (a  +  d)x^  +  (b  +  h)y  +  c-{-k. 

7t  Find  the  sum  of  x^  +  xy  +  y\  ax^—a.ty  +  ay%  And.^by^  +  bxy-{- 
hx\  Ans.  {l+a  +  b)x''-\-{l—a  +  b)xy+{l+a^b)y\ 

8.  Find  the  sum  of  (a  +  b)x  +  {c—d)y—xV2,  {a—b)x  +  {Sc  +  2d)y 
+  bxV2,  2bx  +  3dy—2xV2j  md—Sbx—dy~4:xV2. 

Ans.  {2a—b)x  +  {4c  +  3d)y — 2a;f  2- 

9.  Find  the  sum  of  da'^bf  +  Sa-^b'^'—3a%  —  3(;a"'6''+4/a-' 
fcm-i_a  +  10a3,  and  a'^bP  +  a  +  Sa^b^—2fa-'b'^\ 

Ans.  (6  -  3c)a"'bP  +  (2g''  +  3)a-« J"^*  +  W  +  3a»6'. 

10.  Find  the  sum  of  3-2-'  +  5'',  —8-2-'4-3a"6-'",  and  — 13-5»  + 
4a-2-''  +  carb-^.  Ans.  {4a—5)2-''—12.5''  +  {c-\-3)a"b-^, 

11.  Find  the  sum  .of  9a-'6-V— 76,  lSb—a"b"'+<f—3'2\  and 
3arb"'—?iar^b-^c*  +  3(f—5  •  2^ 

^W5.  (9— A)a-'6-V  +  2a''6'"  +  ll6  +  4<f — 8  .  2\ 

12.  Findthesumof  (a  +  J)f/i"+(2+m)|/y,4y2  +  (a+c)a;2,  3nVy-[- 

{2d—e)x^,  (m  +  n)y^  +  (b  +  2c)^^x,  and  —  2?ii/^  +  12atV. 

-4ws.  [2(a  +  b+d—n)-^  3c— e]  y/x  +  2[3  +  m  +  2{n  +  3a)]  Vy. 


SUBTRACTION. 


(77.)  Subtraction  is  finding  the  simplest  expression  for  the  dif- 
ference between  two  algebraic  quantities. 

CASE    I. 
•  (78.)  When  the  terms  are  entirely  dissimilar. 

RULE. 

Write  the  quantity  to  be  subtracted  after  the  one  from  which  it  is 
to  be  subtra/'ted  with  the  sign  —  between  them. 


4 


30  SUBTRACTION. 

PROBLEM 

1.  Subtract  -f  b  from  a  ;  also,  — h  from  a, 

SOLUTION. 

By  the  rule  we  have  for  the  subtraction  of  +^  from  a,  a— (+6) 
which  is  the  same  as  a— 6.  That  a— (  +  6)  is  the  same  as  a— 6,  may 
be  proved  in  the  following  manner  : 

Since  a-\-h—h  is  equal  to  a;  if  from  a +  6— 6, 

Operation.  +  6  be  taken,  the  remainder  will  be  the  same  as 

a  +  h—h  when  +6  is  taken  from  a.    But,  +6  taken  from 

+  6  a+6— 6  leaves  a—b;  therefore,  +b  taken  from 

a— 6  a  leaves  a— 6;  that  is,  a— (  +  6) =a— 6. 

Bj  the  rule  we  have  for  the  subtraction  of  —6  from  a^  a—{—b) 

which  is  the  same  as  a-f^.    That  a—(—b)\s  the  same  as  a+&,  may 

be  proved  in  the  following  manner : 

Since,  a  +  b—b  is  equal  to  a  ;  if  from  a  +  b—by 

Operation.  —6  be  taken,  the  remainder  will  be  the  same  as 

a+b—b  when  —6  is  taken  from  a.    But  —b  taken  from 

—b  a+b—blesLvesa  +  b  ;  therefore,  —6  taken  from  a 

a  +  6  leaves  a+b;  that  is,  a— (— 6) =a + b. 

PROBLEM 

2.  Subtract  b—c  from  a. 

SOLUTION. 

Since,  a  +  b—b+c—c  is  equal  to  a;  if  from 

Operation.  a  +  6— 6+c— c,  b—c  be  taken,  the  remainder 

a  +  b—b  +  c—c  will  be  the  same  as  when  b—c  is  taken  from  a. 

+  b  —e  But,  b—c  taken  from  a  +  b—b-\-c — c  leaves 

a       —b  +  c  a—b-\-c;  therefore,  b—c  taken  from  a  leaves 

a—b-{-c. 

► 

EXAMPLES. 

1,  Subtract  c->rd  from  a +  6.  -4w5.  a-\-b—c—d, 

2.  Subtract  6— c—e?+e  from  a.  -4w«.  a—b  +  c+d—e. 

3,  Subtract  —  (6+c  +  «?)  from  a.  Ans.  aH-6  +  c  +  <?. 

4.  Subtract  — y— ^fromaj.  Ans.  x+y-^z, 

CASE    II. 
(79.)  When  the  terms  are  similar,  or  partially  similar. 

) 


SUBTRACTION.  31 

RULE. 

Imagine  the  sign  of  the  term  to  he  subtracted  to  he  minus  when  it  is 
-f,  and  PLUS  when  it  is  — ,  and  then  proceed  as  in  addition. 

Remark. — The  reason  of  this  rule  has  been  made  manifest  in  the 
solution  of  problems  1  and  2,  Case  L 

PROBLEM. 

Subtract  6ax—9j/z  from  *lax\-4:yz, 

SOLUTION. 

Operation,  Imagining  Qax—^yz  to  be  —  6aa;  +  9y2  and 

1ax-\-  ^yz  adding  it  to  1ax-\-^yz,  we  have  ax-\-lSyz,  the 

6ax—  9yz  remainder. 

ax+13yz 

EXAMPLES. 

1.  2. 

From  4a  +  36— 2c  +  8c?  From    12xy  +  dy*-l1x^—W2 

Take     a  +  26+  c  +  5d  Take   —5xy  +  1y''—19x^  +  2V2 

Rem.    3a  +  h—Sc  +  Sd  Rem,     11xy—4y''+   2a;'— 5t/2 

3.  4. 

From  28aa:'— 16aV  +  25d'x—lBa*  From  2(a  +  6)  +  3(a— a;) 

Take  ISax'  -{-20a'x^—24a'x—  la*  Take     (a4-&)-3(«— a;) 

Rem.  lOcw;'— 36aV  +  49a'a;—  ea"  Rem.      a  +  6  +6(a— a;). 

5.  From Va;'— /  +  4(a:  +  y)  —  Wa  +  x  subtract  3(a;  +  j^)  —  2(3;' —y^)^ 
+  3(a4-a;)2.  Ans.  3Vx^—y^  +  (x  +  y)—6Va  +  x. 

6.  From  a;' — 2a;y  4-  (x^  -}-  y")  +  (2xy—y'')  subtract  x"  +  2xy—y''  -f  (a;' 
+y')  — 2(2a;y— y'^).  ^ws.  y'— 4a;2/  +  3(2.ry— y'.) 

7.  From  2a'  +  aa;  +  a;' — 1 2a V  +  20aa;' — 4a;'  +  Qa^'x'' — 1  Oaa;'  subtract 
a' — 3aa;  +  2a;'' — 1  Qa'x  + 1 2aa;'' — 1 2a.t' — 4a;'  +  2a  V. 

Ans.  a''  +  4aa;— a;'  +  4a''a;  +  Saa;''  +  2aa;'  +  4aV. 


8.  From  4y'— 4ya;  +  a;'— 2a(a;  +  y)  +  6fa'— a;''— 81^5''— yUake  4a;» 
— 4a;y + y' — 4a(a; +y)—\  OVh'^—y''-^  Wa'—n^, 


Ans.  3y^—Zx''  +  2a{x+y)-\-2Va''—x''-[-2Vb^—y\ 


9.  From   {a  +  hyx^+y''-\-{a  +  c)(a-\-xy  take  (a—hyx'-\-y^  +  c{a 
-\-xy.  Ans.  2h^'¥+f  +  a{a-\-xy, 


B2  SUBTRACTION. 

10*  From  aa;'  +  hyx  +  cy'  take  dx" — hxy  +  ^y'. 

Ans,  {a—d)x^  +  (6  +  ;^)a;y  +  (c— Ar)y'. 

11.  From  a(a;+y)— Jiry+c(a:— y}  take  ^{x-\-y)  +  {a  +  h)xy—1(x 
— y).  ^W5.  (a-4)(a:  +  y)-^(a  +  2%y  +  (c  +  '7)(a:-y). 

12.  From  2x—y-\-{y—2x)—{x—2y)  take  y~2ar— (2y— aj)  +  (a;-|- 
2y).  Ans^y—x, 

13.  FromVa;''— y'— 2(a  +  a?)^  +  3  tate  — 3|/a  +  ar+4(a;*— y'*)^— 1. 

Ans,  Va  +  x—Wx"" — y'  +  4. 

14.  From     2ar(a:+y)*— 3aa;y  +  2a5c     take      —  lYaary  +  llaSc— ar 
\/a;+y.  -4w».  14aa?y— 9a6c  +  3a;(a;4-y)*. 

15.  From  a(a: — yy  ■\-hxy-{-c{a  +  xy  tsikQ  {x—yY—hxy  +  {a  +  c){a 
+xy,  Ans.  (a—l){x—y)'^  +  2hxy—a{a-\-xY, 

16.  From  (a  +  5)(a;  +  y)— (c  +  c?)(a;— y)  +  m  take  (a— &)(a;+y)  +  (c 
—d)[x—y)—n,  Ans,  2h{x  +  y)—2c{x—y)-\-m  +  n. 

1 7.  From  ax^  +  ma;y  +  wa?  +  5  take  sa?'  — ^a:y  +  ga; — c. 

Ans,  {a — sja;"  +  (m  +p)xy  -\-(n— q)x  +  6  +  c. 


18.  From   {a—h)xy—{p-{-q)Vx-\-y—hx''  take  (2p— Sg')  (a:4- y)ir 
— aa;y— (3  +  ;^)ar*.  Ans,  (2a— &)ary— (3^— 2g')y«+y'+3a;^ 

19.  From9a'"a;'— 13  +  20a6V— 46'^ca:'   take   36"'ca;''  +  9a'"a;'— 6  + 
3a6'a;.  ^ris.  mah^x—lh'^cx^—n, 

20.  From    6a*— YaV—Sc-W'  +  Yc?     take    —  15a'6''  +  3a'— 3a''— 
1c~'d\  Ans,  2a*  +  Sa'ft'  +  ^c~'d''  +  1d  +  Za\ 


MULTIPLICATION. 

(80.)  Multiplication  is  finding  an  expression  for  the  product  of 
two  or  more  algebraic  quantities. 

PROPOSITION 

(81.)    1.   When  a  positive  quantity  is  multiplied  by  a  positive 
quantity^  the  product  is  positive,. 

) 


MULTIPLICATION.  38 

DEMONSTRATION. 

Operation.  Let   us   multiply    +2  by    +3. 

+  (+  2)  )            +2\  This  means  that  the  +2  is  to  be 

+  (  +  2)  I    or    +2V    or  +2  taken   positively,  or  additively,  3 

+  (  +  2))            +2 )          +3  times.     Hence,  the  result  is   +6, 

_j_/^g\      _      _j_g    _     _|_g  as  is  shown  by  the  operation. 

The  principle  contained  in  tljis  proposition  is  generally  expressed  in 
the  following 

BULE. 

Plus  (  +  )  multiplied  hy  plus  (  +  )  gives  plus  (+). 

PROPOSITION 

(82.)    2.   When  a  negative  quantity  is  multiplied  hy  a  positive 
quantity^  the  product  is  negative. 

DEMONSTRATION. 

Operation.  Let   us    multiply    —2   by    +3. 

+  (— 2)  )          —  2  )  This  means  that  the  —2  is  to  be 

+  (— 2)V    or   — 2I    or  —2         taken   positively,  or   additively,   3 

+  (—2))         ~2)  +3         times.      Hence,  the  result  is  —6, 

,  /_g\     _     _g    _  _g         as  is  shown  by  the  operation. 

The  principle  contained  in  this  proposition  is  generally  expressed  in 
the  following 

EULE. 
Minus  (— )  multiplied  hy  plus  (  +  )  gives  minus  (— ). 

PROPOSITION 

(83.)  3.  When  a  positive  quantity  is  multiplied  hy  a  negative 
quantity^  the  product  is  negative. 

DEMO  NSTR ATI  ON. 

Operation,  Let   us  multiply   +2   by   —3. 

—  (  +  2)  ')         —2  J  This  means  that  the   +2  is  to  be 

—  (  +  2)>    or  — 2>    or  +2         taken  negatively,  or  subtractively, 

—  (  +  2))         —23  —3         3  times.     Hence,  the  result  is  —6, 
__/_!_ g^     _     _Q     _  _Q         as  is  shown  by  the  operation. 

3 


34  MULTIPLICATION. 

The  expression  —  {-f  2)  is  the  same  as  —2,  because 

Operation.         —(  +  2)  means  that   +2  is  to  be  subtracted.      But 

+  2—2  there  is  nothing  from  which  to  subtract  it.     Let  us, 

+  2  then,  subtract  it  fi'om  nothing,  or  zero.     For  zero,  we 

—  2  write  +2  —  2,  and  taking  +2  from  it  we  have  —2,  as 

is  shown  by  the  operation.     Therefore,  —  (  +  2)  =  —  2. 

The  principle  contained  in  this  proposition  is  generally  expressed 
in  the  following 

RULE. 
Plus  (  +  )  multiplied  hy  minus  (— )  gives  minus  (— ). 

PROPOSITION 

(84.)  4.  When  a  negative  quantity  is  multiplied  hy  a  negative 
quantity,  the  product  is  positive. 

DEMONSTRATION. 

Operatixm.  Let  us  multiply    —2   by   —3. 

This  means  that  the  —2  is  to  be 
or  —2         taken  negatively,  or  subtractively, 
—3         3  times.     Hence,  the  result  is  +6, 
(_6)  +6~  _|_g         as  is  shown  by  the  operation. 

The  expression  —(—2)  is  the  same  as  +2,  because 

Operation.         —(—2)  means  that   —2  is  to  be  subtracted.      But 

+  2—2  there  is  nothing  from  which  to  subtract  it.     Let  us, 

—  2  then,  subtract  it  from  nothing,  or  zero.     For  zero,  we 
+  2                 write  +2  —  2,  and  taking  —  2  from  it  we  have  +  2,  as 

is  shown  by  the  operation.     Therefore,  —(—2)=  +2. 

The  principle  contained  in  this  proposition  is  generally  expressed  in 
the  following 

RULE. 

Minus  { — )  multiplied  hy  minus  (  — )  gives  plus  (  +  ). 

(85.)  The  principles  contained  in  these  four  propositions  may  be 
expressed  by  the  following 

RULE. 

The  multiplication  of  like  signs  gives  plus  (  +  ),  and  the  multipli- 
cation o/"  UNLIKE  signs,  minus  (— ). 


MULTIPLICATION.  85 

PROPOSITION 

(86.)  6.  The  product  of  two  literal  terms  may  he  expressed  hy 
writing  them  in  order^  with  or  without  a  sign  of  multiplication 
between  the  consecutive  terms,  preceded  hy  +  or  —,  according  as  the 
signs  are  like  or  unlike. 

DEMONSTRATION. 

The  truth  of  this  proposition  depends  upon  (19.)  Thus,  a  multiplied 
by  —6  may  be  expressed  by  —axb,  —  a-6,  or  —  «5;  (a  +  h)  multi- 
plied by  (c—d)  may  be  expressed  by  (a-\-b)  x  (c— c?),  {a-\-h)'(c—d), 
or  (a-{-h)  (c—d)  ',  and  (a +  8)  multiplied  by  (4—6)  may  be  ex- 
pressed by  (a  +  3)x(4— 6),  (a  +  3)-(4— 6),  or  (a  +  3)(4— 6). 

PROPOSITION 

(87.)  6.  ^  two  terms,  when  the  exponent  and  sign  of  each  are  not 
considered,  have  a  common  part,  their  product  may  he  expressed  hy  the 
common  part'uffected  hy  the  sum  of  their  exponents,  and  preceded  hy 
-j-,  or  — ,  according  as  the  signs  of  the  terms  are  like  or  unlike. 

DEMONSTR  ATI  ON. 

The  two  terms  +«''  and  —a^,  when  the  exponents  and  signs  are 
not  considered,  have  a  common  part  a  ;  therefore,  we  are  to  prove 
that  the  product  of  -\-a^  and  —a^  is  —  a\  Since,  +(Z^=  4-aa  and  —a' 
^=—aaa,  we  know  by  the  last  proposition  the  product  of  +aa  and 
—aaa  is  —aaaaa.     But,  —aaaaa=:—a^.     Therefore,   -|-a''x— a'= 

Remark. — By  this  proposition,  we  have  -|-2^  x2*=:-j-2^+^=-}-2' 
=4  ;  —a-'  X  —a-'=z  +a-'-'—  +a-'  ;  a-""  x  —  a«=— a-'^+^zz:— a^ 
or  —a. 

GENERAL     RULE. 

(88.)  In  multiplication,  coefficients  are  multiplied,  and  exponents 
are  added. 

CASE     I. 
(89.)  When  both  multiplicand  and  multiplier  are  monomials. 

RULE. 
Multiply  according  to  the  principles  of  the  preceding  propositions 


B6  MULTIPLICATION. 

EXAMPLES. 

1.  Multiply  4:a'b'cdhjSabc'd\  Ans.  l^a'b'c'cf. 

2.  Multiply  12f  ay  by  ^hx,  Ans.  48bxVai/. 

3.  Multiply  5ix'fz*  by  exy'z\  Ans.  3SxYz\ 

4.  Multiply  ISa'ftVy  by  —6abxy^.  Ans.  —  65a^5V/. 

5.  Multiply  —  20a^6?  by  5a'"6V.  Ans.  —  100a'"+^^>"+^c''. 

6.  Multiply  a"*  by  a".  Ans.  a'"+^ 

7.  Multiply  a"*  by  a"",  ^?^s.  a"*~". 

8.  Multiply  a"*"  by  oT.  Ans.  aT"". 

9.  Multiply  a"*^  by  a"".  -4%s.  a-^''^\  . 

10.  Multiply  2a~^,  Ya"**,  auji  —  ^a"  together.  ^tis.  — ^2a~^ 

11.  Multiply  Z'n~\  n~\  and  4*7'  together,  ^tis.  12-f . 

12.  Multiply  -7a-'6*c-*by  Sa'^^-'c.  ^ns.  -21a6-^c-*. 

13.  Multiply  ^aF~\  —  3a''~y,  and  Sa^'ca;  together.' 

Ans.  \W^'"i-^ycx. 

14.  Multiply  — 13a~Vby  -4a~'5~V.  J^5.  52a~'&~''c~\ 

15.  Multiply  a''"^,  arb~\  and  a"^^6  together.         Ans.  a'"6^+\ 

16.  Multiply  {a  +  y)~%n\  {a  +  y)'^H~'m,  and  (a  +  y)  together. 

Ans.  mA^(a  +  y)'"+\ 

17.  Multiply  a^  by  ai».  -4%s.  a. 

18.  Multiply  ai  by  —a" 2,  ^?is.  —  Va. 

19.  Multiply  ai  by  a\.  Ans.  ai. 
20    Multiply  -xT^hyx^.  Ans.  -arzk. 

CASE    II.  1 

(90.)  When  the  multiplicand  is  a  polynomial,  and  the  multiplier 
a  monomial. 

RULE. 

Multiply  each  term  of  the  multiplicand  by  the  multiplier^  connect- 
ing them  by  their  proper  signs. 

PROBLEM.      , 

Multiply  6a  +  45V  -  3d'  by  4a'. 


MULTIPLICATION.  37 

SOLUTION. 

Operation.  •  Multiplying  6a,   +46^c,  and  —Sd^  by 

6a  +   4:b'^c    —dd^  4a'',  respectively,  gives  2 4a^,  +1 6a Vc,  and 

4a''  —12a^d^  which  connected  by  their  proper 

24a'  +  16a=^6'c— 12a''6'        signs  is  24a'  +  16a''6"'c— 12a''<?^ 

EXAMPLE  S. 

1,  Multiplya"— 3a6— 56'by4a''6.     Ans.  4a*6— 12a'6'— 20a=6^ 

2.  Multiply  2a''6'— 5aV  +  9a'^>V  by  3a^bc\ 

Ans.  6a'6V— 15a'5c'  +  2W6V. 


3.  Multiply  2a'— 3c  + 6  by  5c.  Ans.  2a'bc—8bc''  +  5bc. 

4,  Multiply  ax'—bx^  +  cx—d  by  — a;^ 

Ans.  —ax^  +  bx''—cx^+dx^. 

5«  Multiply  5mn  +  8m^—2n^  by  12abn. 

Ans.  60abmn^  -\-36abm'n—24:ahn^. 

6*  Multiply  3ax—5bj/  +  1x7/  by  —*labxy. 

Ans.  — 21a'6a:V  +  35a6'a;y'— 49a6a?y. 

7.  Multiply  —  15a'6  +  3a6'— 126'  by  —bab. 

Ans.  ISa^b^—Ua'^b'  +  eOab*. 

8.  Multiply  a"'af*  +  b"'y"—(fj/"'—d''af'  by  afy*", 

Ans.  a"'x'''+^y''-i-b"*x"'y^"—c"af*y"'^—d"x^"'y". 

9.  Multiply  3x~''—5x'^y~'*+z~'  by  2x~*y"'. 

Ans.  Qx'^y"^  —  1  Ox'^~*y'"~'*  +  2x~*y'^z~\ 

lOi  Multiply  2a~s^  — 7a;~2y*— llcff  by  axy~^c^. 

Ans.  2a\xy~^c%—^axkyc\—Waxy~^c. 

CASE    III. 
(9 1  •)  When  both  the  multiplicand  and  multiplier  are  polynomials 

EULE. 

Multiply  each  term  of  the  multiplicand  by  each  term  of  the  muU 
tiplier,  and  add  the  products. 

PROBLEM. 

Multiply  a'  +  ah-\-b'' hj  a  +  b. 


38  MULTIPLICATION. 

SOLUTION. 

Operation. 

0*+    ab-\-b^  Multiplying  a^-\-ab  +  V  by  a  gives  a' + a'6 

«  +  b  +  a¥;  also,  multiplying  it  by  b  gives  a^J  +  ab^ 

a^+  a'^64-  aft'*  4-*^  which  results  added  produces  a' +  2a''6 

+   a^b^   ab^  +  b^  -\-2ab''  +  b\ 


a»  +  2a'^64-2a6'  +  6' 


EXAMPLES. 

1 .  Multiply  a-\-bhja  +  h,  Ans.  a"  +  2ab  +  6.' 

2.  Multiply  a—bhja—b,  Ans.  a" — 'Hob  +  6». 

3.  Multiply  a  +  bhy  a—b.  Ans,  a'—V. 

4.  Multiply  a"— ab  +  b^  by  a  +  b.  Ans.  a'  -f  b\ 

5.  Multiplya''  +  a6  +  6''by  a-6.  ^w*.  a"— 6^ 

6.  Multiply  a'—a^'b  +  ab^—b^  by  a  +  6.  Ans.  a*  —  b\ 

7.  Multiply  a'  +  a'^  +  aft'  +  i^  by  a-6.  ^ws.  a*-b\ 

8.  Multiply  a*-25'  by  a-6.  Ans.  a''-2ab'-a*b  +  2b\ 

9.  Multiply  »''— 3ar— 7bya?— 2.  ^W5.  rr'  —  Sa;'— a;  +  14. 

10.  Multiply  a'  +  a*  +  a"  by  a""—!.  Ans.  a^—a\ 

11.  Multiply  4a"— 16aa;  + 3ar' by  5a''—2a=a;. 

Ans.  20a'— 88a*a;  + 4  taV- 6a V. 

12.  Multiply  a*  -  2a'b  +  4a'b^  -  Sab'  + 1 66*  by  a  +  2b. 

Ans.  a' +  326'. 

13.  Multiply  I  x^  +  Sax—^a''hj2x^—ax—la\ 

Ans.  5x*  +  iax^  — L<Lla»a?'  +  ia'x + ia\ 

14.  Multiply  16a-''6»-'7a-^6*  +  6a-*6''by8a-='6«-3a-^6S 

Ans.  120a-'b*-101a-"'b'-}-69a~'b'—18a~'b'\ 

15.  Multiply  a*"  +  6^—  2c''  by  2a'«— 36. 

Ans.  2a'""  +2a'"6^— 4a'"c"— 3a'"6— 36?^'+66c'*. 

16.  Multiply  a;~'''  +  3a'«a;~''^— 10a'""a;~P  by  aV  +  5a'^V+^— 2a'"^* 

Ans.  aV""'^  +  8a'^"a;^'^  +  3a""+V~^— 56a''«+V  +  20a*'^a;^^. 

17.  Multiply  Sir +  6,  3a; -f  2,  3a;— 2,  and  3a;— 6  together. 

Ans.  81a;*  — 360ar''  +  144. 


MULTIPLICATION  BY  DETACHED  COEFFICIENTS.  89 

18i  Multiply  Sa—2bj  4a— 36,  4a +  36,  and  3a +  26  together. 

Ans.  144a*— 145a'^6'  + 366*. 

19»  Multiply  x—a,  x—b,  and  x—c  together. 

Ans.  x^~{a  +  b-\-c)x^  +  (ab  +  ac-{-bc)x—abc, 

20.  Multiply  3aV2+7a~*6Hy4afi6"*c*-3a''6-^(r'. 


^  »•  ♦  »»  » 

MULTIPLICATION    BY    DETACHED 
COEFFICIENTS. 

PROBLEM. 
(92.)  To  multiply  by  detached  coeflScients. 

RULE. 

Arrange  the  multiplicand  and  multiplier  according  to  the  ascending 
or  descending  powers  of  a  particular  letter,  and  then  remove  the  letters 
and  multiply  the  coefficients  thus  detached  and  restore  the  letters  ac- 
cording to  the  law  of  exponents  in  each  particular  case. 

DEMONSTRATION. 

Let  us  multiply  x^-^-x^g+xj/^+y^  hj  x—y.     The  terms  in  these 

polynominals  as  they  stand  are  arranged  according  to  the  descending 

powers  of  x  and  the  ascending  powers  of  y. 

Operation.  Removing  the  letters,  we  have  the  coefficients 

1  +  1  +  1+1  l  +  l-f-l  +  l  to  be  multiplied  by  1  —  1.    The 

1  —  1  product,  as  shown  in  the  operation,  is  1  +  0  +  0 

1  +  1  +  1  +  1  +0  —  1.     We  know  that  the  exponent  of  x  in 

—  1  —  1  —  1  —  1  the  first  term  of  the  product  of  the  given  poly- 

1+0  +  0  +  0  —  1  nominal  must  be  *.     Annexing  the  letters  to 

the  coefficients  1+0  +  0  +  0  —  1  according  to 

the  descending  powers  of  x  and  the  ascending  powers  of  y,  we  have, 

lx*-\-0x'g-\-0xy-\-0xy'  —  lg\  or  simply  x'—y\ 

Again,  let  us  multiply  2a^  — 3a6''  +  56^  by  2a^  — 56".      Arranging 


40  DIVISION. 

the  terms  according  to  the  descend- 
Operation.  ing  powers  of  a   or  the  ascending 

2  +  0—  3+   5  powers  of  6,  we  have,  2a''  +  0a'^i— 3 

2  +  0—5  ab'-^5b'   and   2a''  +  0ab-5h\      The 

4  +  0—  6  +  10  product  of  2  +  0— 3 +  5  by  2  +  0  —  5 

_10—   0  +  15  —  25  is  4  +  0  — 16  +  10  — 15  — 25,towhich 

4  +  0  —  16  +  10  +  15  —  25  annexing  the  letters  we  have,  4a^  + 

0a*6- 16  a'b^  + 10  a'b'  +  15ab'-25b% 
or  ^a'-lda'b^  +  lOa'b'  +  Uab'-^BbK 

EXAMPLES. 

1.  Multiply  3a'  +  4aa;— 5a;'' by  2a'— 6aa;  +  4a;'. 

Ans.  6a'— lOa'a;— 22aV  +  46aaJ^— 20aJ*. 

2.  Multiply  a;'-3aj''  +  3a;-l  by  a;''-2a;  +  l. 

Ans,  a;^-5aj'  +  10a;'-10a;''  +  5a;-l. 

3.  Multiply^"— ya  +  ia' by  y'  +  ya—ia'. 

Ans.  y'—aY  +  ia'y—j\a*. 

4.  Multiply  a;*— aj'  +  ic'-rc  +  l  by  ca;'— 6a;' +  aa;. 

Ans.  cx''—(b-{-c)x''  +  {a  +  b  +  c)x^—(a-\-b-\-c)x*  +  {a  +  b-\-c)x'—{a 
+  b)x^-\-ax. 

5.  Multiply     a'-a'bi-a'b'-a'b'  +  a'b^-a'b'  +  a'b'-ab'  +  b'     by 
a—b,  Ans.  a^—b'^. 

6.  Multiply  X*  +  4a;V  4-  6a;y  +  4xf  +  y*  by  x^  +  3x^y  +  3a;y'  +  y\ 
Ans.  x'  +  Ix'y  +  21a;y  +  35a;y  +  35a;y  +  21a;y  +  Ixy'+f. 


^  »•  ^  ••  ^ 


DIVISION. 

(93.)  Division  is  finding  a  factor  of  a  given  quantity  which  mul- 
tiplied into  a  given  factor  will  produce  the  given  quantity,  or  is  find- 
ing how  many  times  one  quantity  is  contained  in  another. 

PROPOSITION 

(94.)  1.  When  a  positive  quantity  is  divided  by  a  positive  quan- 
tity, the  quotient  is  positive. 


DIVISION.  41 

DEMONSTRATION. 

Let  US  divide  +  6  by  +2.  Here  we  seek  a  factor  which  multiplied 
into  +  2  will  produce  +  6.  The  sign  of  the  factor  sought  must  be 
+ ,  or  like  the  sign  of  the  2,  in  order  that  this  factor  multiplied  into 
+  2  shall  produce  a  positive  quantity.  Thus  +6 -7-  + 2  can  not  be 
equal  to  —3,  because  +2  x  — 3  =  — 6  ;  but,  +6-t-  +2=  +  3,  because 

+  2x  +3=:+6. 

Again;  the  division  of  +6  by   +2 

Operation,  ,  ^  j  xi.       +^      t^    ^ 

can  be  represented  thus, .    Factor- 

+  6^+^X+3^  ^  ^  +2 

+  2     +^  ing  the  numerator,  we  have, ; 

and  canceling  the   +2  in  both  terms, 
we  obtain,  +3. 
The  principle  contained  in  this  proposition  is  generally  expressed 
by  the  following 

EULE. 
Plus  (  +  )  divided  by  plus  (  +  )  gives  plus  (+). 

PROPOSITION 

(95»)  2.  When  a  negative  quantity  is  divided  by  a  positive  quan- 
tity, the  quotient  is  negative. 

DEMONSTRATION. 

Let  us  divide  —6  by  +2.  Here  we  seek  a  factor  which  multiplied 
into  +  2  will  produce  —6.  The  sign  of  the  factor  sought  must  be  — , 
or  unlike  the  sign  of  the  2,  in  order  that  this  factor  multiplied  into 
+  2  shall  produce  a  negative  quantity.  Thus,  —  6-T-+2  can  not  be 
equal  to  +3,  because  +2  x  +3=  +6  ;  but,  — 6-^  +  2  =— 3,  because 
—  2x— 3  =  — 6. 

Again ;  the  division  of  —6  by  +2  can 

—  (K 

^        ..  be   represented  thus,  — -.    Factoring  the 

Operatim,  ^  '  +2  ^ 

-6     +^x-3         ^  ^  ,         +2X-3        , 

——=:—— =—3.        numerator,  we   have — ;andcan- 

+  -"      +^  +2 

celing  the   +2  in  both  terms,   we   ob- 
tain, —3. 
The  principle  contained  in  this  proposition  is  generally  expressed 
by  the  following 


42  DIVISION. 

RULE. 
Minm  ( — )  divided  hy  plus  ( -f- )  gives  (  —  ) . 

PROPOSITION 

(96.)  3.  When  a  positive  quantity  is  divided  hy  a  negative  quantity, 
the  quotient  is  negative. 

DEMONSTRATION. 

Let  us  divide  +6  by  —2.  Here  we  seek  a  factor  which  multiplied 
into  —2  will  produce  +6.  The  sign  of  the  factor  sought  must  be—, 
or  like  the  sign  of  the  2,  in  order  that  this  factor  multiplied  into 
— 2  shall  produce  a  positive  quantity.  Thus,  +6-. — 2  can  not  be 
equal  to  +  3,  because  — 2x+3  =— 6;  but,  +6-f-— 2  =  — 3,  because 
—  2x— 3=  +  6. 

Operation,  Again ;  the  division  of  +6  by  —2  can 

I    g  gw  Q  +6 

=  — =  —  3.  be  represented  thus,  — -.     Factoring  the 

o  V 3 

numerator,  we  have,  — ;   and  can- 

—  2i 

celing  the   —2  in  both  terms,  we   ob- 
tain —3. 
The  principle  contained  in  this  proposition  is  generally  expressed 
by  the  following 

RULE. 
Plus  (  +  )  divided  hy  minus  {—)  gives  minus  (— ). 

PROPOSITION 

(97  •)  4.  When  a  negative  quantity  is  divided  hy  a  negative 
quantity^  the  quotient  is  positive. 

DEMONSTRATION. 

Let  us  divide  —6  by  —2.  Here  we  seek  a  factor  which  multiplied 
into  —2  will  produce  —6.  The  sign  of  the  factor  sought  must  be  +, 
or  unlike  the  sign  of  the  2,  in  order  that  this  factor  multiplied  into 
—2  shall  produce  a  negative  quantity.  Thus,  —  6-i — 2  can  not  be 
equal  —3,  because  —  2  x — 3  =+6;  but,  — 6-t-— 2=+3,  because 
--2x+3  =  -6. 


¥ 


DIVISION.  43 

Operation.  Again;  the  division  of  —  6-i — 2  can 

— -  =  — ^ =  +3.  be  represented  thus,  — -.     Factoring  the 

numerator  we   have  — ;  and  can- 
celing the    —2  in    both    terms,  we  ob' 
tain  +3. 
The  principle  contained  in  this  proposition  is  generally  expressed 
by  the  following 

RULE. 
Minus  (— )  divided  hy  minus  (— )  gives  plus  (  +  ). 

(98«)  The  principles  contained  in  these  four  propositions  are  ex- 
pressed by  the  following 


RULE. 
The  division  of  like  signs  gives  plus  (  +  ),  and  of  unlike  signs  (— ). 

Remark. — ^We  may  consider  that  a  positive  quotient  denotes  the 
number  of  times  that  the  divisor  must  be  subtracted 

Operation,  from  the  dividend  to  obtain  zero  for  a  remainder. 
Thus,  we  see  that  +2  must  be  subtracted  three  times 
from  +6  and  —2  three  times  from  —6,  to  obtain 
zero.  Hence  the  quotients  obtained  by  dividing  +6 
by  +2,  and  —6  by  —2  ought  both  to  have  the  same 
sign ;  and,  therefore,  the  quotient  in  each  case 
must  be  +  3. 

We  may  also  consider  that  a  negative  quotient  denotes  the  num- 
ber of  times  the  divisor  must  be  added  to  the  divi- 
dend to  obtain  zero  for  a  remainder.  Thus,  we  see 
that  -f-  2  must  be  added  three  times  to  — •  6,  and  —  2, 
three  times  to  +6,  to  obtain  zero.  Hence,  the 
quotients  obtained  by  dividing  —6  by  +2,  and  +6 
by  —2,  ought  both  to  have  the  same  sign;  and 
therefore,  the  quotient  in  each  case  must  be  —3. 


+6 

-6 

+2 
+4 

—2 
—4 

+  2 
+2 

-2 
—2 

+  2 

—2 

0 

0 

We  may  al 

Operation. 
-6     4-6 

+  2 

-2 

—4 

+  4 

+2 

—  2 

—2 

+  2 

+2 
0 

-2 
0 

44:  DIVISION. 

PROPOSITION 

(99«)  5.  The  quotient  obtained  hy  dividing  one  term  hy  another, 
may  he  expressed  hy  those  factors  of  the  dividend  which  are  not  com- 
mon to  the  divisor, 

DEMONSTRATION. 

Let  us  divide  Qahcd  by  26c.  The  dividend  6ahcdz=2hc  x  Bad,  from 
which  we  see  that  3,  a,  and  d  are  the  factors  of  the  dividend  which 
are  not  common  to  the  divisor.  Therefore,  Sad  is  the  quotient.  Also 
the  quotient  of  axia<m  divided  by  aaxi  is  aa,  because  aaa  of  the  divi- 
dend is  the  same  as  the  divisor,  thus  leaving  a/i  which  is  not  common* 

PROPO  SITION 

(lOO.)  6.  If  two  terms,  when  the  exponent  and  the  sign  of  each 
are  not  considered,  have  a  common  part,  the  quotient  arising  from 
dividing  one  hy  the  other  may  he  expressed  hy  the  common  part  affected 
hy  an  exponent  equal  to  the  exponent  of  the  dividend  minus  the  expo- 
nent of  the  divisor,  and  preceded  hy  +  or  —,  according  as  the  signs 
of  the  dividend  and  divisor  are  like  or  unlike, 

DEMONSTRATION. 

The  two  terms,  —a^  and  +a^,  when  the  exponents  and  signs  are 
not  considered,  have  a  common  part  a.  We  are  now  to  prove  that 
the  quotient  arising  from  dividing  —a^  by  +a'  is  —a^.  Since 
—a''=—aaxiaa,  and  -\-a^=aaa,we  know  by  the  last  proposition  that 
the  quotient  arising  from  dividing  —aaaaa  by  +aaa  is  —aa.  But 
-^aa=:^a^',  therefore  the  proposition  is  proved. 

By  this  proposition,  we  have  —^=a^~^=a^ ;  -^rza"*-";  -^2=a"~^~** 
a 

PROPOSITION 

(lOl.)  1.  Any  factor  may  he  transferred  from  the  denominator  to 
the  numerator  of  a  fraction,  or  from  the  numerator  to  the  denominator, 
hy  changing  the  sign  of  the  exponent. 


DIVISION.  46 


DEMONSTRATION. 


We   have   just    seen    —=«"-'.      But  a^'=— —  ;    therefore, 
;   or,  in  other  words,  a'  has  been  transferred  from  the 


a^     a^cc 


a"         1 

denominator   of  ^-  to  the  numerator  by  changing  the  sign  of  the 

Ob 

exponent. 

Again ;  dividing  both  numerator  and  denominator  of  —^  by  a',  and 

we  have   -j— 5==— -.     But   a^«=a^a-^   therefore,  ^=^i^8» 

or,  in  other  words,  d^  has  been  transferred  from  the  numerator  to  the 
denominator  by  changing  the  sign  of  the  exponent. 

^     , .  .  .  -         a    ab-"^       __j    a       1        a     b'^ 

By  this  proposition,  we  have  j=  -—-=ao    ;  r— ~3iT^  >  j^^y  > 

cd''  a  a"  1       a    a  ' 

From  this,  we  see  that  the  reciprocal  of  a  is  -,  or  a~^ ;  of  a'  is 

—r,  or  «-' ;    of  «-'  is  — r,  or  a\      Hence,  we  may  obtain  the 

a  a 

reciprocal  of  any  quantity  by  merely  changing  the  sign  of  its  ex- 
ponent. 

PROPOSITION 

(1 02,)    8.  Ani/  qvxintity  which  has  zero  for  an  exponent  is  equal 
to  unity. 

DEMONSTRATION. 

If  we  prove  that  a"=:l,  we  shall  prove  the  proposition;  since  a 
may  represent  any  quantity  whatever. 

Weknowthat  -=^=1.    But,  by  Prop.  7,  (101.),^=«'"'=a''. 

CL      CL  €(/ 

d' 
Since,  then,  a°  and  1  are  each  equal  to  -^^  they  are  equal  to  each 

Cb 

other,  that  is,  a^^l. 

GENERAL     RULE. 

(103.)  In  division,  coefficients  are  divided,  and  exponents  sub- 
traded. 


46  DIVISION. 

CASE     I. 
(104.)  When  both  dividend  and  divisor  are  monomials. 

RULE. 
Divide  according  to  the  principles  of  the  preceding  propositions. 

EXAMPLES. 

1*  Divide  abc  by  etc.  Ans.  b. 

2»  Divide  6abc  by  —2a.  Ans.  —36c. 

3«  Divide  —lOxyz  by  5y.  Ans,  —2xz. 

4*  Divide  ISax^  by  Sax.  Ans.  6«. 

5*  Divide  —  28a;V  by  -~4iXy,  Ans.  *lxy'^. 

6«  Divide  a*  by  a\         ^  ^?is.  oT^. 

7.  Divide  a"*  by  a~".  ^ws.  a'''+". 

8.  Divide  ar^  by  a".  ^»s.  «-("+"). 

9«  Divide, a""*  by  a"".  ^W5.  a'"^". 

ca^* 
lOt  Divide  ca**  by  da~\  Ans.  —j-. 

11.  Divide  —Sa'^'b^hj  —^M^c\  Ans.  — . 

12.  Divide  6  (a  +  by  by  4  (a+6)-\  Ans.  -,  ^^.,. 

13.  Divide  (a  +  a:)Xa4-y)-'  by  (a4-a?)->  +  y)-'. 

^7j«.  (a  +  a;)"(a+y)*. 

14.  Divide  UOa'b'cd'  by  30a'6'(f .  ^/w?.  Ba^b'cd. 

15.  Divide  15a"Vy  by  3a"*a;'>'*.  ^»«.  Sa'^aJ^^T^. 

16.  Divide  —^Sa^'b""  by  6a^5^.  ^n«.  — Sa'^ft"^. 

2  1  12  Q^ 

17.  Divide  ea^c?"^  bv  3a~3<^^  -4iw. -p. 

18.  Divide   12a~^dic~*  by  3a8c?~«c^.  Ans.  —. 

''  arc 

19.  Divide  (a  +  a?)"*  by  6(a4-a;)^.  -^n« 


6(a  +  a;)' 


k 


DIVISION.  47 

20.  Divide  {a-\-b)^{x  +  y)~^  by  (a  +  by^{x-\-yp\ 

{a  +  by 

CASE     II. 

(105.)  When  the  dividend  is  a  polynomial,  and  the  divisor  a 
monomial. 

RULE. 

Divide  each  term  of  the  dividend  by  the  divisor^  and  connect  the 
quotients  by  their  proper  signs. 

PROBLEM. 

Divide  6a'6*-8a''6'c?'4-4a*6'c  by  2a'b\ 

SOLUTION. 

Dividing  Qa^b\  —Sa'b'd^  and 

Operaticm,  +4a*6''c    respectively,    by    2a'' J*, 

2a%)  ''6a''b'Sa''b'd'  +  4a*b^c  gives  Bab%  4bd\  and  +  2a'c,  which, 

2ai)^  —4bd^     -\-2a^c  connected  by  their  proper  signs,  is 

Sab''—4bd''  +  2a'c. 

EXAMPLES. 

1 .  Divide  1 2a'x  -\-  4aic'  —  1 6a  by  4a.  Ans,  Sax + a?' — 4. 

2.  Divide  12ay-16ay +  20ay-28ay  by  -4ay. 

Ans.  —3y^-\-4ay^—5a^y-\-'7a*, 

3.  Divide  15a'5c— 20acy'  +  6ccP  by  —5abc. 

4y»     rf" 

Ans.  —  3a  +  -r 7-. 

b       ah 

4.  Divide  3;''+'— a?'*+'  +  af+'— af»+*  by  x\       Ans.  x—x^-^x^—x\ 

5.  Divide  a"+'a;— a"^a;— a"+'a;— a»+*a;  by  a". 

Ans.  ax—a^x—a^x—a*x. 

6.  Divide  aaf  -{■ax''-^'  +ax*^''  ^ax"^  by  a;". 

Ans.  a-\-ax-\-a^-\-a3^, 

7.  Divide  6(a;+y)'-8(a;  +  y)»  +  4a^(a?+y)by  2(a:4-y). 

Ans.  Z{x-\-yY-^{x-\-y)-\-2a\ 

8.  Divide  5(a  +  6)»-10(a+6)»  +  15(a  +  6)  by  -5{a  +  b). 

Ans.   -(a  +  6)''  +  2(a4-6)-3. 


48  DIVISION. 

9.  Divide  asf^^  +hx'^^—Car-^+dx''  by  x^^, 

Ans.  ax^+bx''—cx*+dx^^-^. 

10.  Divide  ia'x^—^ax^  +  Sab^xhj  faV. 

CASE    III. 
(106,)  When  both  dividend  and  divisor  are  polynomials. 

RULE. 

1.  Arrange  both  dividend  and  divisor  according  to  the  ascending 
or  descending  powers  of  the  same  letter  in  both. 

2.  Divide  the  first  term  of  the  dividend  by  the  first  term  of  the 
divisor  ;  the  result  will  be  the  first  term  of  the  quotient,  by  which 
multiply  all  the  terms  in  the  divisor,  arid  subtract  the  product  from 
the  dividend. 

3.  Then  to  the  remainder  annex  as  many  of  the  remaining  terms  of 
the  dividend  as  are  necessary,  and  find  the  next  term  of  the  quotient 
as  before,  and  so  on, 

PROBLEM 

1.  Divide  6aV  +  a*— 4a'a:  +  a;*— 4aa;'  by  x'  +  a''—2ax, 

SOLUTION. 

Arranging  the  terms  according  to  the  descending  powers  of  a,  we 
have 

a''—2ax  +  x')a*~4a'x+6a''x^—4:ax'-{-x\a'—2ax+a^ 
a*  —  2a^x-[-  aV 

—  2a'a;  +  5aV— 4aa;' 

—  2a^a;  +  4a  V  —  2aa;' 

a^x^—2ax'-\-x* 
aV— 2aa;'+a;* 

0 

ANOTHER     SOLUTION. 

Arranging  the  t^rms  according  to  the  descending  powers  of  x,  we 
have 


DIVISION. 

x^—2xa+a' 


— 2x^a  +  5x^a^ — 4txa^ 

x^a*—2xa^+a* 
x^a*—2xa^+a* 
0 

PROBLEM 

2.  Divide  2a»*— 6a''*6'*+6a"6''*-26'''  by  a"— 5*. 

SOLUTION. 

2ci^"— 2a''"6" 

2a"6''*— 26'» 
_ 

PRO  B  LEM 

8.  Divide  a;'— (a4-6+c)a;''+(a64-ac  +  6c)a:— a6c  by  x—e, 

SOLUTION. 

a;*-—  (a + 6  4-  c)x^  +  {ab-^ac  +  bc)x — dbc^  - 


x—c 


x^—(a  +  b)x-i-ab 


—  (a + 6)a;'^  +  (a6 +ac-\-  bc)x 

—  (a  +  6)a;^  +        (ac  +  6c)a; 

o5a;      —      abc 
abx       —       abc 

0 

EXAMPLES. 

1«  Divide  a*  4- 2a6  + 6' by  a +  6.  Ans.  a+b. 

2.  Divide  a'—2ab-\-b^  by  a— 6.  .4ws.  a—b. 

3.  Divide  a?*+4a;V  +  6a;y  +  4a;y'+y*  by  a;'  +  2a;y  +  y'. 

^W5.  aj'-f  2a:y4-y'. 
4 


50  DIVISION, 

4*  Divide  ic* — ^x^y  +  ^x'y^ — 4a;y' + y*  by  x—y. 

Ans.  x^—Sx'y  +  Sxy^—y*, 
5t  Divide  x^—y*  by  x—y,  Ans.  x^  +  x^y  +  x^y^  +  xy^  +  y*. 

6.  Divide  ic'  +  y^  by  a;+y.  ^^.  ar*— a;'y  +  a;y— ary'+y*. 

7.  Divide  a:^— 9a;' +  2 7a;— 27  by  a;— 3.  Ans.  a;'— 6a;  +  9. 

8.  Divide  12a;*— 192  by  3a;— 6.  Ans.  4a;' +  8a;'' +  16a; +  32. 

9.  Divide  dx' — 6y'  by  2a;' — 2y\  Ans,  Sx*  +  3a;y  +  Sy\ 
lOi  Divide  a;'  +  5a;'y4- 5a;y'+y^  by  a;''  +  4a;y+y^         Ans.  x+y. 

11.  Divide  a'-Za'¥ -[-^a'h'-h'  by  a'-3a'6  +  3a6'-6^ 

Ans.  «'4-3a'6+3a6'  +  6'c 

12.  Divide  a^-h'  by  a^  +  a%^-db^+h\  Ans.  a-b, 

13.  Divide  ia;'+a;'  +  ia;+f  by  ia;  +  l.  Ans.  a;'  +  3, 

14.  Divide  x^-\-y*  by  a;+y, 

Ans.  x^—x^y-\-xy^—y^  +  -    ^ 


x+y 

15.  Divide  a;"'+'  +  a;"'y+a;y'"  +  y'"+' by  a;"*+y"*.  Ans.  x+y. 

16.  Divide  a;*"  +  x^Y^-^-y*"  by  ar"*  -f  a;"y~ + y'". 

17.  Divide  a'^6"— 4a"''+^'6'*— 27a'*+'^V*  +  42a'»+^'5*»    by  a^^h^ 
—  '7a"-'b'\  Ans.  a'«+3a'^'6"— ea''*-'^'*. 

18.  Divide  a**- a;"  by  a— a;. 

a'-^-V— af* 


^%s.  a  '^^ + ar-''x+ a'^ V + • 


a—x 


19.  Divide    ar*  — (a+6+rf)a;'  +  (a(/  +  ^'<i^+c)a;-cc?   by  a;'— (a +6) 
a;  +  c.  ^9w.  a;— rf. 

20.  Divide  -a%*-}-15a^'b'-4:8a'*b'—20a"b'hj  10a'b^—a*b. 

Ans.  a'b'—5a'b*—2a'b\ 

tl.  Divide  4c*—9b^c'  +  6b'c-b*  by  20"- 36c +  6'. 

22.  'Divide  faj"*— 4a;* 4-V a;'- 4Jia;'- 3^33.^27  by  ia;'— a;4-3. 

^n5.  fa;'— 5a?'  +  Ja;  +  9. 

2f8r  Divide  —l+aVby  —l-^-an.  Ans.  l+a«+aV. 


DIVISION.  51 

21.  Divide   a'd'-^da'cd'  +  Sac'd'-c'd'+a'c'd^-ac'd^    by    a'd'- 
2acd^  +  c^d''  +  cu;''d,  Ans.  ad—cd, 

25.  Divide  l-6s»  +  2lz*  by  i+2z+3z\  Ans.  1  -60  +  9z\ 

26.  Divide  —  2a-V4-l'7a-V-5a;'— 24aV  hy  2a-'x'—3ax\    . 

Am.  — a-V+'7a-V  +  8aV. 

27.  Divide  a''^''"5'^c— a''"H^'6'-^c'' + a-'^J-'c"* + a'"^6'H- V— a'^H^"-* 
6 V^'  +  ^p+^c^H^'  by  a-'*6-^'  +  bc"^'. 

Ans.  a^'^'^^'^'c— a''»+''^'6V  +  6^c". 

28.  Divide  4(3a^*5-i4.7a"^V5T*5)c*_3(3aV5~V-4.7a''*V^«)c?-« 


DIVISION    BY    DETACHED 
COEFFICIENTS. 

PROBLEM. 
(107»)  To  divide  by  means  of  detached  coefficients. 

RULE. 

Arrange  the  terms  of  the  divisor  and  dividend  according  to  the 
ascending  or  descending  powers  of  a  letter  common  to  both ;  then, 
omitting  the  letters,  write  the  coefficients  with  their  respective  signs, 
supplying  the  coefficients  of  the  absent  terms  with  zeroes.  Proceeding 
as  usual  in  division,  the  result  will  be  the  coefficients  in  the  quotient, 
to  which  annexing  the  letters  acc(yrding  to  the  law  in  each  particular 
case,  will  give  the  complete  quotient. 

PROBLEM . 

Divide  6a*— 96  by  3a— 6. 

SOLUTION. 

Arranging  the   coefficients   as  directed   and  dividing,  we  obtain 


5i  SYNTHETIC  DIVISION. 


Operation. 

3-6 

for  the  coefficients  in  the  quo- 

6 +  04-0  +  0—9612  +  4  +  8  +  16 

tient,  +  2,  +4,  +8,  and   +16. 

6-12 

An  inspection  of  the  problem 

12+   0 

shows  that  the  first  term  of  the 

12—24 

divisor     should     contain      a'. 

24-  0 

Therefore,  commencing  with  a' 

24-48 

and  inserting   the  descending 

48-96 

powers  of  a,  we  have  for  the 

48-96 

complete  quotient,  2a^  +  4a'  + 

0 

8a +  16. 

EXAMPLES. 

1.  Divide  a;*— 3aa;'— 8aV  +  18a'a;— 8a*  by  a;''  +  2aa;— 2a'. 

Ans.  a;'— 5aa;  +  4a'. 

2.  Divide  3/  +  3a;y''-4a;V-4a;' by  a;  +  y.        Ans.  _4a;»  +  3y''. 

3.  Divide  10a*— 27a''iC  +  34aV— 18aa;'— 8a;*  by  2a''— 3aa;  +  4aj^ 

Ans.  5a'—6ax—2x*. 

4.  Divide  a»  +  4a'— 8a*-26a'  +  35a»  +  21a-28  by  a''  +  5a  +  4. 

Ans.  a*— a^— '7a'  +  14a— 7. 

5.  Divide  oT+^+xf'—cf'y—y'^'  by  x'^  +  y^  Ans.  x—y. 


SYNTHETIC    DIVISION. 

PROBLEM. 
(108.)  To  divide  by  sjmthetic  division. 

RULE.* 

1 ,  Divide  the  divisor  and  dividend  by  the  coefficient  of  the  first 
term  in  the  divisor,  which  will  make  the  leading  coefficient  of  the 
divisor  unity,  and  the  first  term  of  the  quotient  will  be  identical  with 
that  of  the  dividend. 

2.  Change  all  the  signs  of  the  terms  in  the  divisor,  except  the  first, 
and  multiply  all  the  terms  so  changed  by  the  term  in  the  quotient,  and 

*  This  rule  is  due  to  Mr.  "W.  G.  Homer  of  Bath,  England. 


SYNTHETIC  DIVISION.  53 

place  the  prodiicts  successively/  under  the  corresponding  terms  of  the 
dividend,  in  a  diagonal  column,  beginning  at  the  upper  line, 

3.  Add  the  results  in  the  second  column,  which  will  give  the  second 
term  of  the  quotient ;  and  multiply  the  changed  terms  in  the  divisor 
by  this  result,  placing  the  products  in  a  diagonal  series  as  before. 

4,  Add  the  results  in  the  third  column,  which  will  give  the  next 
term  in  the  quotient,  and  multiply  the  chanyed  terms  in  the  divisor 
by  this  term  in  the  quotient,  placing  the  products  as  before. 

5*  2%is  process  continued  until  the  results  become  0,  or  until  the 
quotient  is  determined  as  far  as  necessary,  will  give  the  same  series  of 
terms  as  the  usual  mode  of  division  when  carried  to  an  equivalent 
extent. 

Remark. — ^In  synthetic  division,  as  in  division  by  detached  coefficients,  it  is 
customary  to  omit  the  letters. 

PBOBLE  M. 

Divide  x'-5x^  +  15x^'-24:X^  +  2lx''-lSx  +  5  by  «*-.2jf'  +  4«"— 
2aj+l. 

SOLUTION. 

Since,  in  tlds  problem  the  coeflBcient  of  the  first  tenn  of  the  divisor 
is  unity,  Part  1st  of  the  rule  is  unnecessary. 

(Part  2d  of  the  Rule.)  Omitting  the  letters,  arrange  the  dividend 
horizontally,  and  the  divisor  vertically,  changing  the  signs  of  all  its 
terms  except  the  first. 

Operation.  Then  multiply  the  changed 

1  1—5  +  15  —  24  +  27—13+5  terms  in  the  divisor  by  1,  and 
+  2     +2—6  +  10  placetheproduct, +2,  —  4, +2, 

— 4  —4  +  12—20  and  —1  diagonally  under  —6, 


l_5  +  15_24  +  27-13+5 

+  2- 

6  +  10 

— 

4  +  12  — 

20 

+  2- 

6  +  10 

— 

1+  3-5 

1-3  + 

5+   0  + 

0+  0+0 

+  2  +  2—  6  +  10  +15,  —24,  and  +27  respect- 

—  1 —  1+  3—5         ively. 

1  —  3+   5+   0+   0+  0+0 

(Part  3d  of  the  Rule.)  Adding  the  results  in  the  second  column, 
gives  —3  for  the  second  term  of  the  quotient,  which  multiplying  into 
the  changed  terms  of  the  divisor,  gives— 6,  +12,  —6,  and  +3,  which 
must  be  placed  diagonally  under  +15,  —24,  +27,  and  —13  respect- 
ively. 

(Part  4^  Adding  the  terms  in  the  third  column  gives  +  5  for  the 
third  term  of  the  quotient,  which  multiplying  into  the  changed  terms 
of  the  divisor,  gives  + 10,  —20,  +10,  and  —5,  which  must  be  placed 


64  SYNTHETIC  DIVISION. 

diagonally  under  —24,  +27,  —13,  and  +5  respectively.  Here  the 
process  terminates,  since  the  sum  of  each  of  the  remaining  columns 
equals  0. 

Restoring  the  letters  according  to  the  law  of  the  case,  we  obtain  for 
the  quotient  sought  x^—3x-\-5, 

EXAMPLES. 

1.  Divide  a'— 6a*a;  +  10aV— 10aV  +  6aa;*— «'  by  a'— 2aa;+a?'. 

Ans.  a^—Sa^x  +  Sax^^x^, 

2.  Divide  a'— 3aV  +  3aV— aj"  by  a^—Sa''x+Sax^—x\ 

Ans,  a^+3a''a;+3aaj'+aj'. 

3.  Divide  x''—y''  by  x—i/. 

Ans.  a;'+a;V+^y+^y+^y+^/+y°' 

4.  jyiyidQ9x'—4:6x^-\-95x^  +  l50xhjx''—4x—5. 

Ans.  9a;*— lOaj'+Sa;'— 30a?. 

5.  Divide  25a;''-a;*-2a;''— 8a;»  by  5a;'-4a;^ 

Ans.  6a;'+4a;'+3a;+2. 

6.  Divide  a»— 6a*a!  +  10aV-10aV+6aa?*-ar'*  by  a'-3a''a;4-3aa;» 
— a;»,  Ans,  a^—2ax+x\ 


CHAPTER   III 


THEOREMS  AND   FACTORING 


THEOREM     I. 

(109.)  The  square  of  the  sum  of  two  quantities  is  equal  to  the 
square  of  the  first,  plus  twice  the  product  of  the  first  hy  the  second, 
plus  the  square  of  the  second. 

DEMONSTRATION. 

Let  a-\-h  represent  the  sum  of  two  quantities.  Squaring  it,  or 
multiplying  it  by  itself,  we  have  a^  +  2ah-\-h^,  or  (a  +  6)"=a'  +  2a6 
+  6^ 


Square  2a'' +  36' 


PROBLEM. 


SOLUTION. 


By  the  theorem,  we  have  (2a')''+2(2a'')(36')  4-(36')^  which,  after 
the  operations  indicated  are  performed,  becomes  4a*  +  12a''6^  +  96^ 


EXAMPLES 


1.  Square  x-\-y. 
2t  Square  2x  +  y, 

3.  Square  ^x^  +  ^y\ 

4.  Square  x^-{-y^. 

5t  Square  3a''  +  4a6'. 
6t  Square  ^c^-^dn^, 
7»  Square  a~^  +  6~\ 
8.  Square  a^  +  h\. 


Ans,  x^  +  ^xy+y"*, 

Ans.  4:X^  +  4:Xy-\-y^. 

Ans,  9x'  +  24:xY  + 16/. 

Ans.  ic'  +  2a;y+y*. 

Ans.  9a*  +  24a'b'  +  iea^h\ 

Ans.^G*  +  c^dn''  +  d^n*. 

Ans,  a-^  +  2a-'b-'  +  b-\ 

Ans.  ay  +  2a|-H  +  6. 


5d     •  THEOREMS  AND  FACTORING. 

9.  Square  2a-i  +  Sa-ib-h  Ans.  -  +  -ii-+-A-. 

a     ab-k     abt 

10.  Square  3x-ii/i  +  2a:iy-f .  Ans,  M  + 12  +  ^. 

1 1  •  Square  ^a^bx~^ + |a~2 6~3a;~^.      ^ws.  — =-  -\ h  ^  7  * 7 1  • 

12.  Square  -i-  +  -T^.  -4w«.  4a-»5  +  — 7-+ia*6-^ 


THEOREM    II. 

(1 10.)  77ie  square  of  the  difference  of  two  quantities  is  equal  to 
the  square  of  the  first,  minus  twice  the  product  of  the  first  by  the 
second  J  plus  the  square  of  the  second, 

DEMONSTRATION. 

Let  a—b  represent  the  difference  of  two  quantities.  Squaring  it, 
or  multiplying  it  by  itself  we  have  a'— 2a6  +  6^  or  {a—by=a''—2ab 
+  b\ 

PROBLEM. 

Square  5a' —36*. 

SOLUTION. 

By  the  theorem,  we  have  (5a')»— 2(5a')(36*)  +  (36*)»,  which,  after 
the  operations  indicated  are  performed,  becomes  25a^— 30a'6*+96', 

EXAMPLES. 

1.  Square  jr—y.  Ans.  x^—2xt/+y^. 

2.  Square  ia;"— 20:2^.  Ans,  lx*—2xY-\-4:xY' 

3.  Square  im^—iTi*.  Ans,  }m'—lm^n*  +  ^n\ 

4.  Square  5ay 3 _6oj»y^^  ^^s,  25aV^— 6<^«V  +  36aV^. 

5.  Square  ab~^  —  2 la-'6'.  Ans.  -rr 1 r- . 

^  b^       a         a* 

6.  Square  2a*6'—Ta-'6-^  Ans,  4aV— 28a6-j-49a-°6-*. 

o       Q  1    'i       8  *)      4 

7.  Square  5a''6'— 4a* 63.  Ans.  25a*'6*— 40a'*  6^ +16a2  63. 


THEOREMS  AND  FACTORING.  67 

8.  Square  Sa-'fe""— 4a-'6- 


Ans.  9a-*h-^ — 24a-'5-'  + 1 6ar^h-\ 

5  _5         -J  3                        ,         25a''     ^        496' 
9.  Square  5a^b  ^-1a  H\  Ans.  -^, 70  4—^. 

10.  Square  Sx  +^-4a;— y".  Ans.  9x'^'^+'-24.xr  +  ^' 

_2±1  »-|-L 

11.  Square  2ar("»-^V  ^   -Sic'^+'y     ^  . 

w-fft+3    _f?iZL  -»n-«-3    »«i« 

12.  Square  Ja;     «     y     2   _4a;  y  4  , 


-4W5. 


THEOREM    III. 

(111.)  The  product  of  the  sum  and  difference  of  two  quantities 
is  equal  to  the  difference  of  their  squares. 

DEMONSTRATION. 

Let  a-\-h  and  a—b  represent  respectively  the  sum  and  difference  of 
two  quantities.  Multiplying  a-{-b  by  a—b,  we  have  a^—b'^,  or 
{a-\-h)(a-b)=.a^-b\ 

PROBLEM. 

Find  the  product  of  2a'' +  36'  by  2a'— 36'. 

SOLUTION. 

By  the  theorem,  we  have  (2a')'— (36')",  which,  after  the  operations 
indicated  are  performed,  becomes  4a*— 96°. 

E  X  AMPLES. 

1.  Find  the  product  oim-\-n  and  m—n.  Ans.  m^—n*. 

2.  Find  the  product  of  ia'  +  i6  and  K— l6.         Ans.  la'—^^. 

3.  Find  the  product  of  Ya'6' +  6c(?  and  1a''b^—6cd. 

Ans.  49a16*— 36c'<f'. 

4.  Find  the  product  of  a~^  +  6"*  and  a~^— 6"^        Ans.  —5 — tj- 


58  THEOREMS  AND  FACTORING. 

5.  Find  the  product  of  Sab~^  +  ^a~^b  and  dab~^—^a~% 

6.  Find  the  product  of  mi  +  wi  and  mi— wi.  ^ws.  m— w. 

7.  Find  the  product  of  2ia;i  +  3iyi  and  2ia;i— Siyi. 

-4ws.  2a;— 3y. 

8.  Find  the  product  of  3a'6'  +  2ai6f  and  3a'6'— 2ai6f. 

9.  Find  the  product  of  ^a^b-^ +^ar'b'  and  ^a'b-'-^ar'b\ 

4a'       9b' 
Ans. 


9b'     16a" 

lOt  Find  the  product  of  ic^+y"*  and  a:*^—y'".  Ans.  a;'"*— y'" 

11,  Find  the  product  of  ^a'^'  +  ^a'^'b  and  2a'^'— Ja'^^J. 

Am.  W-+'-    ^"^      ^ 


25 

wt— 3     3— ■>»  3—^    m— 3  *»^3     3— *» 

12.  Find  the  product  of  fa  2  6  a    +fa  2  6  2    and  |a  3   6  2 

3—m  »t— 3  4a"^3     96"*— 3 

— |a  262.  Ans.  — rr— o  — 7-;r-o' 

THEOREM     IV. 
(112.)   The  difference  of  two  quantities  is  divisible  by  the  differ- 
ence of  the  same  roots  of  the  quantities. 

DEMONSTRATION. 
Let  xn—y»  be  a  general  expression  for  the  difference  of  two  quan- 

m     \  t     \  m  t 

tities,  and  {x^)'^—{y~*)'^,  or  x^—y~>  be  a  general  expression  for  the 
dijfference  of  the  same  roots  of  the  two  quantities. 

m  t  m  t 

We  are  to  prove  that  xn—y7  is  divisible  by  x^—y~». 

m  t  m  t^ 

Dividing  x'^—y'^  by  xm—yr,^  we  obtain  for  the 

m       m     t_  t 

1st  remainder     x'^~^y~»—y». 

m        2"*    2'  * 

2d  "  X^~  ~Tny~rr — y~s. 

m       3wi      3t  t 

8d  "  x'n      7^y17 — yT. 

m       Ttn,         rt  t 

yth         "  X^     rtTyTB — yT, 


THEOREMS  AND   FACTORING.  59 

rm         m  rt         t  m      m    J^  t  t 

Since,  "5^—"^,  and  ^=7,  this  remainder   z=zx^~'^y—y^=:afy»^ 

t  L      1      L 

y..    Because  by  Prop.  8,  (102.)  x°=l,  we  have  xy»—y=y 

t 

— y7=0. 

m  t  m  t 

Hence,  x^—y^  is  divisible  by  x^—y^,  because,  after  obtaining  r 
terms  in  the  quotient,  the  remainder  equals  zero.     The  form  of  the 

mm  m      2"*      '  "*     S^"      2^ 

quotient,  as  we   see   by  division,   is  x'^~'^-\-x'^    ^y7'3-^xn~  r»y  ra 

m     t  __2^  t        t_ 

,, ,  ,,  Xmy  «         r«    -j-  y «       r»  , 

m  2 

Q^n 0/ J  mm  m      2"*     '  »»       3"*      2'  2*'*     <        3*_ 

Hence,  —^ j=zX'n~'^  -{-  X'n~  '^y'Vi  -\-  X'^       nTy"^ Xmy»        n 

Xm yrt 

m      t  _  2t  t       t 

+  Xmyl       «   +y.      «      (A.) 

When  — =tt,  and  -=z^,  and  r=:w,  we  have ^=:a;'*~^  +  a;'*~V + 

iP'-y arV-'-hary—'  +  y—'.     (B.) 

PROBLEM 

1.  Divide  x^—y^  by  x—y. 

SOLUTION  . 

x^—y^ 

Making,  in  formula  (B.),  %= 5,  we  have  =zx^-\-x^y-^x'^y*-{' 

x — y 

xy^'  +  y*. 

PROBLEM 

2.  Divide  a^—b^  by  (a^)^—  {bi)  5,  or  a^^—b'^^. 


8OLUTIO  N 


*^      w-^*^ 


Making,  in  formula  (A.),  — =|,  -=|,  and  r=5,  we  get  — =1^ 

/fr  O  fit) 

a.      ,  7 

t  a  ^  b  ^        SL  -9 

and  — =T^5  •    Also,  putting  x=za  and  y =&,  we  have — — =a2-i  tt 

-f-ai»6>s+6»5=as  +a^  ^b^  ^  +a^b^  ^  +a^  ^b^  +b^  K 


JO  THEOKEMS  AND  FACTORING. 

EXAMPLES. 

1  •  Divide  x^—a^hj  x— a,  Ans,  x^  +  ax-\-  a^, 

2.  Divide  a^—h"  by  a—h.  Ans,  a*  +  a'6  +  a'6"  +  a6H6*. 

3.  Divide  a'— 6"  hy  a—h, 

Ans.  a"  +  a'h  +  a'6'  4-  a^h^  +  «&*  +  h\ 

4.  Divide  a'—li'  hy  a—h. 

Ans.  a'  +  a'h  +  a*b^  +  a'b'  +  a%'  +  ah''  +  h\ 

5t  Divide  a^—b^  hj  a—h. 

Ans.  a'  +  a'b  +  a'b'  +  a*b'  +  a'h'  +  a'h'  +  ah'  +  h\ 

Li  3.  i         JL  3. 

6.  Divide  x^—y^  by  x^—y^,  Ans.  x^  +xy^  +x^y-\-y^, 

2.  2. 

7.  Divide  x^—y^  by  x^  — y^. 

a.      A3       £A      sfi.      1 

8.  Divide  x^—y"  by  x^—y^. 

Ans.  x^  +a;  3  y6  4.3:2^3  ^x^y^  +x^y  ^  +y  ^  , 
9i  Divide  x^—yhj  x\  —yh  Ans.  xz  +  xyi  +  ici^ri'  +  y^. 

2.  JI  _2_  i 

10.  Divide  m^  — ti^  by  m^  ^  —n^ . 

A  1.0.    L  J_    L  2     3  _4_    i  _2_    i.  3. 


Ans.  m'^  +^2  1^8  -|-m2»y4  4-m^2/8+m2»y2 +^212^8  ^^4 

11.  Divide  a;'— y"'  by  x^—y-^. 

2.  3.    _3.  3    _3  _£ 

^ws.  iC* +a;2y  * -\-x*y  ^ +y  *. 

12.  Divide  a~^— a;^  by  a~2j_a;3s. 

_L6  12     _4_  ?_    _8_  4-    -12  J.J8 

Ans.    a     254.^     2Sa.35_|_a     25x^5^a     ZS^^S^x^^. 

THEOREM     V. 

(1 1 3»)  The  difference  of  two  quantities  is  divisible  by  the  sum  of 
the  same  roots  of  the  quantities,  when  the  index  of  the  root  is  even, 

DEMONSTRATION. 

m  t 

Let  x'^—y'^  be  a  general  expression  for  tbe  difference  of  two  quan- 

tities,  and  x^n^r'^  \y  3)  r^  or  xm+yra  he  a  general  expression  for  the 
sum  of  the  same  roots  of  the  two  quantities.     We  are  to  prove  that 

xn—yi  is  divisible  by  xm-\-yr$^  when  r  is  an  even  number. 


THEOREMS  AND  FACTORING. 


Dividing  ic«—y»  by  xm-^-yra^  we  obtain  for  the 

m        m        t  t 

1st  remainder    —xn ~  myVa  —yl 


2d 

u 

m       2»»      2«             * 
Xn        rn  y  ra   — y  » 

8d 

(( 

|»_3i^    _3*_            *_ 
—Xn       rmy  r»  — y« 

4th 

u 

TO      4ff»     4<             t 
Xn       my  n  —y. 

tn        rm      rt  t 

rth  remainder,       x^    '^yVs—yl  when  r  is  an  even  number. 
This  remainder  =0  as  was  shown  in  the  Dem.  of  Theorem  4th. 

m  t  m  t 

Hence,  a;"^— y »  is  divisible  by  icm  +  y^,  when  r  is  an  even  number,  be- 
cause after  obtaining  r  terms  in  the  quotient,  the  remainder  equals 
zero. 

The  form  of  the  quotient,  as  we  see  by  division,  is,  a?^~m— ajn""^ 

t            TO      3n»      2<            »»       4'»      3<  ^     t       2t  J_        * 

yr»-{-a;n        m  y  ra  — Xn        ntWrj-}-, Xmy  a         ra  — ya       ra, 

—  i. 

/j;„ yj  TO       TO  TO        a***     _^ 

Hence,  when  r  is  an  even  number,  -^^ j=zxn~~m—xn~'^yr*-{- 

Xm-{-  yra 

TO       3»»     2'  "»      4"*       3'  3"*    J_       4*  2*»     J 3* 

ajn"       fny~rr  —  iCn~~rny~ra    ■\- X  m  y  a~   ra  —  X  my  a      n, 

m     t       2t      t       t 

+a:^yT     7ry'7~7i,     (A.) 

When  — =u:  -=w  ;  and  r=u,  r  being  an  even  number,  we  have 
n  s 

^ =a;«-i  _a;«-V  +  x^'-y —x'^-y  + — ic'y'^' + xy"*-* 

x—y 

-y^^   (B.) 

PROBLEM 

1.  Dividea'— 6"by  a  +  ft. 

SOLUTION. 

Making,  in  formula  (B.),  w=6,  we  have 
a-\-o 


62  THEOREMS  AND  FACTORING. 

PR  OBLE2i 

2.  Divide  a^—h"  bya«+6«. 


OLUTION. 


Making,  in  formula  (A.),  -=5, -=5,   and  r=6,  we  get  — =- 


and  — =-. 
rs     6 


Also,  putting  x=a,  and  y=&,  we  have 

=a     6— a     6  66  .^a     6  6     —a     ^b^+a     f^  b  ^  —b  ^  z=z 


fO 


a6+66 

a  6  —a  3""66  ^a^hi —a^h^  j^f^^if^- _^- ^ 

EXAMPLES. 

1 .  Divide  a* — 6*  by  a  +  6.      ^  */4ws.  a' — a^b  4-  a6* — 6' 

2.  Divide  a®— 6^  by  a +  6. 

Ans.  a'-a'b  +  a'b^-a'b'  +  a'b^-a'b'  +  ab'-b' 

3.  Divide  a— 6  by  ai  +  6i.  -4?i5.  a^—aibi  +  albi—bi, 

4.  Dmde  a^—b"  by  a*  +6*.         Ans.  a  *  —a^b^  -^a^b^—b  * 

5.       3.  j_s.       5.  ^        f.  3.      a. 

5.  Divide  m^—n^  by  m*  +w*.     ^w*.  m  *~— m^w*  +m*w2_^4 

ji  i         i       ajL 

6.  Divide  x^  —y'  by  a; + y * .  Ans,  x^ — ic'y *  +  ary ^  —y  ♦ 


7.  Divide  «'— y°  by  a:^  +y. 

J  A  2  1 

Ans,  x^—x^y  +  xy^  +  x^y^-\-x^y*—y^, 

7.  3. 

8.  Divide  x''—y^  by  a;«  +3^*. 

4JL  2J_    3_  JL5_    3.  7.    S.  2JL    3  1    JJL  1    a  aJL 

^ws.  x^  — a;  *  y8  +«  8  y4  _a;2y8  .i.^.  a  yz  —x^y  «  +a;82/4  _y  s  , 

9.  Divide  x^—y^  by  a;^  +y"TF"^ 

-4ws.  x^—x^y^'^  4-  a?6  y *  — y  *  ^. 

10.  Divide  x-^—y~^  by  a;-'+y-'. 

^W5.  ar"— a;-V~'+a:-V~'— y~'- 

—3.  —3.  -_3.         —fi. 

11.  Divide  a;*— y~' by  a;  +  y  *.        Ans.  x^—x^y  ^ +xy  ^—y  *. 

12.  Divide  x^—y~^  by  a;~^^  +  y~*^ 

T_  2. 5  _3_    ft_  i 5_  _l_    _10.  _a5. 

Jins.   x^—x^y  *2_{_./^io^  2i_-j.sy   i4_j_,^i02^  2^— y  '*2. 


THEOEEMS  AND  FACTOEING.  6S 

X 

THEOREM    VI. 

(114.)  The  sum  of  two  quantities  is  divisible  by  the  sum  of 
the  swme  odd  roots  of  the  quantities. 

DEMONSTRATION. 

Let  «» +y"r  be  a  general  expression  for  the  sum  of  two  quantities, 

and  (a;»)^+  \y^p,  0YX~^+y^  be  a  general  expression  for  the  sum 
of  the  same  roots  of  the  two  quantities. 

We  are  to  prove  that  xn-\-y»  is  divisible  by  a;m+yr.,  when ris an 
odd  number. 

Dividing  x^-^y'^  by  xrn-{-yr»^  we  obtain  for  the 

m       m    m  t 

1st  remainder      —x'^~ ^y~» + yT. 

m      2"*      2*  * 

2d        "  x'^~~T^y~^+yT, 

3d  "  —Xn       my  r,   +y.. 

4th        "  x^     "^y^  4-  yT. 


m      rm       tt  t 

rth  remainder      —x^    Wy'TT-^ys  when  r  is  an  odd  number. 

TO  t_ 

This  remainder  is   obviously  equal   to   zero.     Hence,  a;n+y«   is 

divisible  by  a;"»+y«,  when  r  is  an  odd  number,  because  after  obtain- 
ing r  terms  in  the  quotient,  the  remainder  is  zero. 

mm  m       Qm     t 

The  form  of  the  quotient,  as  we  see  by  division,  is  x'^~'m—x'^~~^yr» 

m       3m    2'                                                     2jn     t        3<            m      t  2<           _<  t 

-j-  X^       rrTyrs  •\- -\- X  m  y  t         ra  X^nV  a  ra    -\- y  a  f», 

m  t_ 

Xn-{-ya  ^_^  ^_2^    i. 

Hence,  when  r  is  an  odd  number  —^ -=xn    m—x*     r7iyrs-{- 

x^-\-y^ 

w*      3CT      2*                                                 2»n      «        3t  m     t        2<  *        * 

Xn       m  y  ra    ......  ...-j-iC»^y«         •"*   Xmy  a         *■*    "j"  V  •       '*.        (-"••) 

When  — =^*,  -=w,  and  r=u,  r  being  an  odd  number,  — 


n        '  s       '  '  "  «  +y 

PROBLEM 

1.  Dividea*  +  6'by  a  +  6. 


&4c  THEOREMS  AND  FACTORDiTO. 


SOLUTION, 


Main  Tig,  in  formula  (B.),  u=5,  we  have r-=a*— a"6+a*6'— 


ab'  +  b\ 


PROBLEM 


2.  Divide  a^  +  b^  hy  a^+b^, 

SOLUTION. 

Making,  in  formula  (A.),  — =8,  -=8,  and  r=:Sj  we  get  — =-, 

and  — =-. 
rs     3 

Also,  puting  x=aj  and  y=^bj  we  have 
a3  +  63 

EXAMPLES. 

1.  Divide  a'  +  6'  by  a  +  b,  Ans,  a^—ab-\-b\ 

2.  Divide  a"4-6"  by  a'  +  6'.         Ans.  a^-a'b^-\-a'b*-a'b'  +  b\ 

8.  Dividea'  +  6'by  a  +  6. 

Ans.  a'-a'b-ha*b'-a'b'-{-a'b'-ab'  +  b\ 

L 


4.  Divide  a  +  bhj  a^  +b~' 


4.         3.1         2.2.         iS.         1. 

^Tis.  a*— a^ft* +a*6^ +a^6« +65. 


5.  Divide  a«  +  6'  by  o*  +  6«. 


-<4w«.  a  '^  —a  ^  b^  +a  ^  b  ^  —a^o^  +6  *  . 
6.  Divide  a"  +  6'  by  a^  +  6. 


.4w«.  a^  —a^  b  +  a  «  6"— a«6'  +  6*. 


7.  Divide  a'  +  6*  by  a  +  6*.     ^»5.  a*— a'6^  ^a^'b^—ab'^  +  b^^. 


8.  Divide  a  +  6'  by  a^'  +  bK 


A  J.2  2. A  J.,6         ,8 


Ans.  a^  —a^b^  -\-a^b^—a^b^  -\-b^, 

9.  Divide  a-'-\-b-'  by  a-'  +  b-\ 

Ans.  a-*-a-'b-'  +  a-^b-'-a-'b-'  +  b~\ 


THEOREMS  AND  FACTORING.  65 


10.  Divide  a^ -i-b^  hj  a'^^ +b^\ 


_8_  _6_    _3_  _4_    _3_  ^_    _9_  J 


11.  Divide   a^  +  6  Hya^^+i  ^^^ 

_8_  _6_  4_  _4_ 8_  _2_    _JL2.  15. 

THEOREM    VII. 

(11 5.)  The  sum  of  the  squares  of  two  quantities,  plus  twice 
the  product  of  the  quantities,  is  equal  to  the  square  of  the  sum  of  the 
quantities, 

DEMONSTRATION. 

Let  a  and  b  represent  any  two  quantities,  then  a^  +  b^+2ab,  or  a' 
+  2a6  +  6^  represents  the  sum  of  their  squares  plus  twice  their  pro- 
duct. 

We  are  to  prove  that  a'  +  6'  +  2ab,  or  a'  +  2a6  +  6" = (a  +  b)\  By 
Theorem  1,  we  have  (a  +  by=a''  +  2ab-\-b^\  ,-.  a^  +  2ab  +  b\  or 
a^  +  b''  +  2ab=(a  +  b)\ 

PROBLEM 

1.  Find  the  factors  of  oi^+4x+4, 

SOLUTION. 

Since  x^  +  4:X  +  4,  or  x^-[-4:  +  4:X=x^  +  2^  +  2'2x,  it  represents  the 
sum  of  the  squares  of  x  and  2,  plus  twice  their  product.  Hence,  by 
the  Theorem,  we  have  a;' +  2' +  2-2a;,  or  x''  +  4x  +  4=(x  +  2y=:{x  +  2) 
{x  +  2). 

EXAMPLES. 

1 ,  Find  the  factors  of  a;" + y'  +  2xy.  Ans.  (a: +y)  (^  4-  y). 

2«  Find  the  factors  ofa^  +  2ay  +  t/*,  Ans.  (a-{-y)(a-\-y). 

3.  Find  the  factors  of  a;'  +  10a;  +  25.  Ans.  (x  +  5){x  +  6). 

4.  Find  the  factors  of  ar'  + 24a; +  144.  Ans.  (a;  +  12)(a;  +  12). 

5.  Find  the  factors  of  m''  + 114m +  3249. 

Ans.  (m  +  57)(m  +  57). 

6*  Find  the  factors  x  +  2xiyi + y.  Ans.  {xi  +  yi) (ar^  +  yi). 


W  THEOREMS  AJSTD  FACTORING. 

7.  Find  the  factors  of  4x^-\-4:X+l.  Ans.  {2x  +  l)(2a;  + 1). 

8.  Find  the  factors  of  169a;'*  +  26a: +1.    Ans.  {lSx  +  l)(lSx  +  l). 

9.  Find  the  factors  of  25ar-''  +  60a;-Vi  +  36yf . 

Ans,  (5x-^  +  6yi)(6ar'  +  6yi). 

10.  Find  the  factors  of  a;+a+2aiari^.       Ans.  (xi+ai){xi  +  ai). 

11.  Find  the  factors  of  xi+a'x^  +  2ax. 

3.  J         3.  A 

Ans.  {x^  +ax^){x^  +ax^). 

12.  Find  the  factors  of  9xi  +  49a;2y~s  +  42a;y'"^.     • 

Ans.  {3x^-\-1x*y-'^)(dx*+1x*y-^). 

THEOREM     VIII. 

(116.)  The  sum  of  the  squares  of  two  quantities,  minus  twice 
the  product  of  the  quantities^  is  equal  to  the  square  of  the  difference  of 
the  quantities. 

DEMONSTRATION. 

Let  a  and  b  represent  any  two  quantities,  then  a^-\-b^—2db,  or 
a^—2ab  +  h'^  represents  the  sum  of  their  squares,  minus  twice  their 
product.    We  are  to  prove  that  a^  +  b^— 2abj  or  a^ — 2ab  -\-b^=(a— by. 

By  Theorem  U.  we  have  {a—by=a^-2ab  +  b^ ;  .  • .  a''^2ab  +  b% 
OT  a^  +  b''-2ab={a-by. 

PROBLEM. 

Find  the  factors  of  9a; +  4yi—12a;iyi. 

SOLUTION. 

Since  9a;  +  4yi— 12a;i3^=(3a;i)'  +  (2yi)''— 2-(3a;i)(2?/i),  it  repre- 
sents the  sum  of  the  squares  of  two  quantities,  minus  twice  their  pro- 
duct Hence,  by  the  Theorem,  we  have  9a;  +  4yi— 12a;iyi,  or  9a;— 
12a;iyi  +  4yi=:(3a;i— 2yi)''=(3a;i— 2yi)(3a;i— 2yi). 

EXAMPLES. 

1.  Find  the  factors  of  a;''—2a;y  +  2/^  Ans.  {x—y){x—y), 

2.  Find  the  factors  of  m'*— 4 w^?^  +  4?l^      Ans.  (m—2n){m—2n), 

2x 

3.  Find  the  factors  of  x^-\-^ — -.  Ans.  (a;— i)(a;— i). 


THEOREMS  AND   FACTORING.  67 

4.  Pmd  the  factors  oia^—2a^x^+x^,    Ans.  {d^ —x''){a^ —x^y 

5.  Find  the  factors  ofa'—4aV  + 4/.    Am,  {a^ —2y^)((fi —^y""), 

6.  Fmd  the  factors  of  9a;'  +  9y— 18a;V^. 

Ans.  (3a;*— 32/i)(3a;*— 3yi). 

7.  Find  the  factors  of  :^a;'— 3a;^  +  4.     Ans,  (|a;2_2)(f a;^— 2). 

16 

8.  Find  the  factors  of  16a;f — 16a;i2r'  +  4y-'. 

Ans.  (4a;i— 2y-')(4a;i— 23r'). 

9.  Find  the  factors  of  3a;f +3y-f— Garfy-i. 

Ans.  (34a;f-3i2r-4)  (3Ja;f-3iy-i). 

10.  Fmd  the  factors  of  a;'""  +  y'''»— 2a;'"y''. 

Ans,  {f—y%(if^—'f)- 

11.  Find  the  factors  of  4a;*'»-16a;""y^  +  16y". 

Ans.  (2a;'^'«— 42^)(2a;'""— 4yl). 

12.  Find  the  factors  of  a;^y~T— 2+ a;  «y». 

Ans.  {x~2^y    2«— a;   2'»y2«)(a;2n— a;   2«y2#). 

THEOREM     IX. 

(117,)  The  difference  of  the  squares  of  two  quantities  is  equal  to 
the  product  of  the  sum  and  difference  of  the  quantities. 

DEMONSTRATION. 

Let  a  and  b  represent  any  two  quantities,  then  a^—b'  represents  the 
difference  of  their  squares. 

We  are  to  prove  that  a'^—¥={a  +  b){a—b). 

By  Theorem  m.,  we  have  (a  +  6)(a— 6)=a'— 6"),  .*.  a'— 6"= 
{a  +  b){a—b). 

PROBLEM. 

Find  the  factors  of  a— 6. 

SOLUTION. 

Since  a— 6=(ai)'— (H)",  it  represents  the  difference  of  the  squares 
of  the  quantities  ai  and  bi. 

Hence,  by  the  Theorem,  we  have  a— 6=(ai  +  H)(ai— &i). 


68  THEOREMS  AND  FACTORINa. 

EXAMPLES. 

1,  Find  the  factors  of  a;'— y^  Ans.  (x-\-y){x—y) 

2,  Find  the  factors  of  ic*—y'.  Ans.  (a;*  +  y')(a;*— y') 
3*  Find  the  factors  of  a?'— y.                  Ans,  (a^+y¥)(a:f--y|-) 

4.  Fmd  the  factors  of  a;»-y^  Ans.  (a;t +yf)(a;f-y|) 

5.  Find  the  factors  of  ar-*-2r'.  Am.  (^  +  -)(^'-2rO 

6.  Find  the  factors  of  9a;''— 4y*.  Ans.  (3a;+2y')(3a?— 2y') 

7.  Find  the  factors  of  2a;— 2y. 

Ans.  L(2a;)i  +  (27y)i](2ia;i-2iyi) 

8.  Find  the  factors  of  3'a;^— 6y«. 

Ans.   {3M  +  52yTV)(3ia;6_5^yTV) 

9i  Find  the  factors  of  4:X^—y~^.     Ans.  l^x^  +  -A  (2x^ r) 

10.  Find  the  factors  of  16ar-''-25y.  Ans.   (i+Syi)  (--5yA, 

11.  Find  the  factors  of  4af*— G^r". 

2a;2+3y~^j  (2a;2— 3y  ^), 

2m  2m         2* 

12.  Find  the  factors  of  a?"""^- 4a;~Jry    T. 

Xn      y»    '     Xn       y*    ', 
THEOREM    X. 
(11 8.)  The  expression  x^—yT=\x^—y7»)\xn  m-{-xn    myr,-{- 

m     3m      2*  2m      t       3*  m       t      2<  _£ t\ 

Xn       my  r«   + -{-X  m  y  9      n    ■\-XTny  »       ra  ■\-y  a     r$  f  ^    whlch, 

Wi       t 

when  — ,  -,  and  r  respectively  equals  u,  becomes  a;"— y''=(a;— y) 

n     s  ' 

(ar-^+a.«-«y+a;«-y  + +x^y'^''  +  xy'^^+ir'\ 

DEMONSTRATION. 

The  truth  of  this  Theorem  depends  on  the  truth  of  Theorem  IV.,  and 
the  fact  that  the  divisor  multiplied  by  the  quotient  equals  the 
dividend. 


THEOEEMS  AND  FACTOBINQt,  $& 

PROBLEM 

1.  Fmd  the  factors  of  x^—y^ 

SOLUTION. 

In  this  M=5,  and  making  r=6,  we  have,  by  the  Theorem,  a;*— y* 

Other  factors  may  be  obtained  by  assigning  other  positive  integral 
values  to  r, 

PROBLEM 

2.  Find  the  factors  of  xl—yh 

SOLUTION. 

In  this  example,  — =-,  and  -=-,  and  we  are  at  liberty  to  make 

r  any  positive  integer.    Let  r  then  equal  6 ;  whence,  — — ^»  ^^^ 

— =-—.     Hence,  by  the  Theorem,  we  have  x* —y^  =  [x'^^ —y^'^j 
rs     15 

(3  92  34  32  8\ 

a;5  4-a;2  0yTj^a;ToyT5>|_^2  0ys  _|_yTsj,     Other  factors  may  be  ob- 
tained by  assigning  different  positive  integral  values  to  r. 

EXAMPLES. 

1.  Find  the  factors  of  a'— a;'. 

Ans,  (a—x){a''  +  ax-\-x''),  or  [a^—x^)  [J  +  x^), 

2«  Find  the  factors  of  a^—x*. 

Ans.  (a—x){a-{-x),  or  \a^—x^)  [a^  +  ax^ +aix  +  x^) , 

3.  Find  the  factors  of  a^—x\ 

Ans.  (a-x){a' -ha'x  +  a'x' -{-a^x' -{-ax*+x%  or  {a'-x'){a'+x'). 

it  Find  the  factors  of  d'—x\ 

Ans,  {a—x)  {a"  +  a^x  +  d'x'  +  a"a;' +a''a;*  +  ax^ + x\) 

5*  Find  the  factors  of  a'— a;^ 
Ans.{a'-x')(a'  +  a'x^+x'\oY  {J-x^)  {J +  a'x^ +  Jx'  +  x^) 

6.  Find  the  factors  of  a'^—x'". 

Ans.    (a-a:)(a''+a'a;+aV  +  aV+aV  +  aV+aV  +  aV  + 
ax'+x'). 


70  THEOREMS  AND  FACTORING. 

7.  Find  the  factors  of  a'"— a;". 

Ans.  (a''-ar')(a«  +  aV4-aV  +  aV+a;'),  or  {a'-x'')(a''+x'), 

8t  Find  the  factors  of  m—w. 

Ans.  {mi—ni){m^  +  mhii-\-nf)j  or  ^mi—ni;){mi+ni), 

9»  Find  the  factors  oix^—y*. 

Ans.  {x''-y){x'+x'y  +  xY+y%  or  {x'-f){x'+y^). 

10.  Find  the  factors  of  x~^—yz. 

Ans.  {ar'-y^){x-'-{-x-''y^+(xr'y^'\-y\  or  (ar-''-yf)(^+yf j. 

11,  Find  the  factors  o{af'—y\ 

3  3 

12*  Find  the  factors  of  x^—y  *, 

Ans.  I XI s— )ia;i5+ + + + — ). 


THEOREM    XI. 
(119.)  The  expression  x'^—y'^=  \x^  +  y m j  [x^ ~m—xn     ^yV, -f 

Xn        m  y  rs    , ..,.,,    X  rn  y  a         r»   -^X^ny  s         ra   — y  a       ra  ) 

when  r  is  an  even  number,  which,  when  — ,  -,  and  r  respectively 

equals  u,  becomes  x''—y''={x+y){x'^^—x'*~^y  +  x'^^y^— — 

xy-'-^-xy^'-y^K 

DEMONSTRATION. 

The  truth  of  this  Theorem  depends  on  the  truth  of  Theorem  V., 
and  the  fact  that  the  divisor  multiplied  by  the  quotient  equals  the 
dividend. 

PROBLEM 

1.  Find  the  factors  of  x^—y^. 

SOLUTION. 

In  this  example  w=6,  and  making  /=6,  we  have,  by  the  Theorem, 
a;"— y°=z(a?  +  2/)(a;'— a;*y4-a;y— icy  +  a?/— /).  Making  r= 2,  we  have 
x^^y^^{x^j^f){x'^f). 


THEOREMS  AND  FACTORING.  71 

Other  factors  may  be  obtained  by  assuming  r  equal  to  other  even 
numbers. 

PROBLEM 

2.  Find  the  factors  of  a;*  *  — y^. 

SOLUTION. 

In  this  example,  — =-,  and  -=-,  and  we  are  at  liberty  to  make 

/i         Tc  So 

r    any   positive    even  number.       Let  r  then   equal   6  ;     whence, 

— =-,  and  — =-.    Hence,  by  the  Theorem,  we  have, 
rn     8  rs      9  '    -^  '  * 

Other  factors  may  be  obtained  by  assuming  r  to  be  equal  to  other 
even  numbers. 

EXAMPLES. 

1.  Find  the  factors  of  a*—b*, 

Ans.  (a  +  b)(a'^a'h  +  ab'-b%  or  (a'  +  6»)(a"~6'),&c. 

2.  Find  the  factors  of  a—b. 

Ans.  (ai  +  bl){ai—aUi  +  aibi—bl),  &c. 

3.  Find  the  factors  of  a'—b^^ 

Ans.   \a  +  b)  (a—b),  or   [a^  +  b^)  [a^—a^b^  +  ab^—aH 

-\-ah^—b^),  &c. 

4.  Find  the  factors  of  a'—b'. 

Ans.   (a^  +  b)  {a^-ah  +  ab'^-^ah'  +  ah^-b'),  &o. 

5.  Find  the  factors  of  ar*—b*. 

\a      /\a'      a"      a         / 

6.  Fmd  the  factors  of  a-'—b-'\ 

/11\/1         1         1         i        1        1\„ 

7.  Find  the  factors  of  aJ  +  b\ 

Ans.   {J-b')  {J—ah'  +  ah*-b'),  &c 

8.  Find  the  factors  of  16a^-b'\ 

Ans.  (4a'  +  6^)(8a*'-— 4aV  +  2a'6«-6").  &c. 


72  THEOREMS  AND  FACTOEINa. 

9.  Find  the  factors  of  «*'"—«"». 

Ans.  (a"*  +  a;"»)(a''"*— a""a;»'»  +  a'"a;*"— a;"*),  <fcc. 
10.  Find  the  factors  of  a'—bK 

Ans,   \ai+b^l  [a  ^  —aabi-^-aAba^bi),  <fec. 

11  •  Find  the  factors  of  a^—b-h 

12t  Find  the  factors  of  a    «  —  6  s  . 

1       IL 
Am.  /--+bi\{—^ — -L+_^_6-f\,  &C. 
an  '  ^a  »      an       a  »  ' 

THEOREM     XII. 

(120.)    The     expression    x^+y~^=  \x^  +  y^ j  \x^~~^—x7k    m yn 

m  _  3Wt     2<  a"*    i._3f  ^     <        2*  * ^\ 

•\-Xn       my  n  — , -{-iJ^rn  y »      r»  —  Xny  t       r«    -\-y  »      rij  . 

when   r  is   an   odd   number,   and,   when  — ,  -,  and  r  respectively 

equals  w,  it  becomes 

x-  +  y-={x^-y)  (a:'^^-a;'-V  +  x«-y-a;-y  + +x^y^ 


■xy'^'+y'^K 


DEMONSTRATION. 


The  truth  of  this  Theorem  depends  on  Theorem  YI.,  and  the  fact 
that  the  divisor  multiplied  by  the  quotient  equals  the  dividend. 

PROBLEM 

1.  Find  the  factors  of  a^  +  bK 

^  SOLUTION. 

In  this  example  m=5,  and  making  r=5,  we  have  by  the  Theorem, 
a'  +  6'=(a  +  5)(a*-a'6  +  a='6'»-a6H6*).      Making  r=3,    we    have 
a3  +  63 j  ^^-3  _a353  4.5-3  . 

Other  factors  may  be  obtained  by  assuming  r  equal  to  other  odd 
numbers. 

PROBLEM 

2.  Find  the  factors  of  x%+y^. 


THEOREMS  AND  FACTORING.  .  78 

SOLUTION. 

In  this  example  — =-,  and  -=-,  and -we  are  at  liberty  to  make  r 
^      «      4  s     3      .  '' 

any  positive  odd  number.     Let  r  then  equal  5,  whence  — ==^»  ^^d 
rs      15.     Hence,  by  the  Theorem,  we  have  x* +y^=\x^^ +y^^) 

/    3  _9_    _2_  3       ^_  _3_    5  JL\ 

Other  factors  may  be  obtained  by  assuming  r  to  be  equal  to  other 
positive  odd  numbers. 

EXAMPLES. 

1 ,  Find  the  factors  of  a^  +  a;\         Aiis.  (a  -\-x){a*—ax-\-  a;'),  &c 

2.  Find  the  factors  of  a^-\-x*. 

/     3  3\/    12_  9.    3  6     £  3     ^  J^\ 

3«  Find  the  factors  of  a'  +a\ 

Ans.  (ai-\-xi)  (a^^'^—ah^-{-b^^J,  <feo. 
!•  Find  the  factors  of  a'+a\ 

Ans,   [a^  +x^)  [a^  —a^x^+a^x^  —a^x  «  +a; «  j ,  &0. 

5.  Find  the  factors  of  a~'-}-a;~^ 

Ans.   (-  +  -)(— +  — ),  &c. 

\a     xj\  x^     ax      a;V 

6.  Find  the  factors  of  a-'  +  a;". 

Am.  (l+^^)(i,_4  +  ---+^'Uc 
\a        /\  a*      or      a       a         / 

32      1 

7.  Find  the  factors  of  -^  H — 5. 

a        X 

/2       1\/16        8         4  2  1\    ^ 

Va'*     ar/\  a®      a"a;     aV       aV       «*/ 

8.  Find  the  factors  oi  a^  -\-x^, 

Ans,  [a^  +  x'^^^)  [a^—a^x^^  -h  x^^J ,  <fec. 

9.  Find  the  factors  of  a^'^+a:""; 

Ans,  (a'"  +  a;'")(a*'"— a'"'aj'"'4-a''"*a;*''— a'"a;«"4-aj«''),  &c 

^       -i         • 
10*  Find  the  factors  of  a«  +a;» . 


a  3»  -f  aJT^j  \a¥^ — oz^x  s*  4-  a;  3*  / ,  <fcc. 


74  THEOEEMS  AKD  FACTORING, 


11.  Find  the  factors  oi  a^+x  K 


2  4  3 

^  x^^  a:2s      a;25     a;25      3.25/ 


X 

li»  Find  the  factors  of  an+32x-\ 


3m 


THEOREM     XIII. 

(121.)  The  trinomial  aj'""+aa:'"  +  6  =  (af"+c)  (a:"+rf),  when 
e+d=a  and  cd=b  ;  and  aj''"*  +  aaf*— 6=(a;'"  +  c)(a;"»— <?),  when  c—d 
=za  and  cd=b  ;  and  a;^'^— aa;"*— 6=(.r'"+c)(a;"'— fl?)  when  c?— c=:a 
and  cd=b ;  and  a;'''"— aa;'"  +  6=(a;"'— c)(a;"*— rf)  when  c4-c?=za  and 
cd=h. 

DEMONSTRATION, 

By  multiplying   x'^+c  by  oT  +  d,   we   have   a;'"*+(c  +  <?)af'  +  cc?, 
which  is  equal  to  x'^."*  +  ax^  +  by  when  c-{-d=a  and  cd^=&. 
In  the  same  way  the  other  forms  may  be  proved  to  be  true. 

PROBLEM. 

Find  the  factors  of  a;"— a;— 30. 

S  OLUTION. 

In  this  example  m=l.  Let  us  now  seet  two  factors  of  30,  one 
plus  and  the  other  minus^  which  have  a  difference  of  1,  the  minus 
factor  being  numerically  the  greater.  It  is  apparent  that  these  factors 
are  —6  and  +5;  hence,  according  to  the  third  form,  we  have 
aj'—a-— 30=(a;  +  6)(a;— 6). 

EXAMPLES. 

1;  Find  the  factors  of  x^'  +  Sx+e.  Ans.  (a;+2)(a;+3). 

2.  Find  the  factors  of  a;'' +  8a; +  16.  Ans.  (a;+3)(a;  +  6). 

3.  Find  the  factorsof  a;'  +  8a;  +  '7.  Ans.  (a;  +  l)(ar+'7). 

4.  Find  the  factors  of  a;'  +  4a;— 32.  Ans.  (a;4-8)(a:— 4). 

5.  Find  the  factors  of  x^—^-{-20.  A71S.  (a:— 4)(a;— 5). 

6.  Find  the  factors  of  a;'— 5a?— 66.  Ans.  (a;  +  6)(ar— 11). 

7.  Find  the  factors  of  aj«— a;'-6.  Ans.  (a;' +  2) (a:' -3). 


THEOKEMS  AND   FACTORmG.  75 

8.  Find  the  factors  of  a;'+a;  +  f.  Ans,  («  +  |)(a;4-i.) 

9.  Find  the  factors  of  a;="»— a;"*— 72.   ^       Ans,  (ar  +  8)(af'—9.) 

10.  Find  the  factors  of  a;"— fa;— f|.  Ans.  (x  +  i){x—^). 

11.  Find  the  factors  of  x^  +  (a  +  b)x  +  ab.         Ans.  (a;  +  a)(a;4-6). 

r  r 

12.  Find  the  factors  of  a;T  +  3a;'i^  +  2. 

Ans.  (ara* +l)(a;2» +2). 

PEOBLEM     A. 
(122.)  To  resolve  a  monomial  into  factors. 

RULE. 

Assume  any  monomial  as  one  of  the/actors^  and  divide  the  given 
monomial  by  it  to  obtain  the  other  factor. 

PROBLEM. 

Resolve  a'  into  two  factors. 

SOLUTION. 

Assume  that  a'  is  one  of  the  factors,  then  —T=ia~^  is  the  other. 

EXAMPLES. 

1.  Resolve  a*  into  factors.         Ans.  a^-a^^  or  a^'a~\  or  6*^-,  &o. 

0 

2.  Resolve  a~^b^  into  &ctors.  Ans.  — T*a'6',  <fec. 

a* 

3.  Resolve  6a  into  factors.  Ans.  3a-2a°,  &c. 
4(  Resolve  95^  into  factors.                                 Ans.  Sx°'3b^,  &c. 

5.  Resolve  12a6  into  factors.  Ans.  Sax'4:bx~^,  &c. 

6.  Resolve  20a'a;-''  into  factors.  Ans.  4a"ar*-5a~*a;~*,  <fec. 

7.  Resolve  (a +  6)'  into  factors.  Ans.  (a  +  6)(a4-&)',  <fec. 

8.  Resolve  (a-\-b  +  c)*  into  fiictors. 

9.  Resolve  15ay  into  factors.  Ans.  Sa'y^'^,  <fec. 


76  THEOREMS  AND  FACTORING. 

10*  Resolve  21ai  into  factors.  Ans,  Sa'-^-,  <feo. 

at 

PROBLEM    B. 

(l!23«)  To  resolve  a  polynomial  into  factors  one  of  which  shall 
be  a  monomial. 

RULE. 

Assume  any  monomial  as  one  of  the  factors  and  divide  the  given 
polynomial  by  it  to  obtain  the  other  factor, 

PROB  LEM 

1.  Resolve  a'  +  ic'  into  factors  one  of  which  shall  be  a  monomial. 

SOLUTION. 

a^'  +  a;'  a;*  . 

Assume  a  to  be  one  of  the  factors,  then =«  H is  the  other. 

a  a 

PROBLEM 

2.  Find  the  fectors  of  3aa;' +  66a;*  +  9ca:*. 

SOLUTION. 

Assume  3a;'  to  be  one  of  the  factors,  then  (3aa;''  +  66a;*+9ca;*)-T-3a;" 
=« + 26a;  +  3ca;^     Hence,  3aa;' + 66a;*  +  9ca;' = 3a;' (a + 26a;  +  3ca;'). 

Scholium. — In  order  to  obtain  the  simplest  factors  of  a  polynomial, 
assume  as  one  of  the  factors,  the  greatest  monomial  that  will  divide 
the  given  polynomial  and  give  a  quotient  containing  neither  a  frac- 
tion, nor  a  negative  exponent. 

EXAMPLES. 

1 ,  Find  the  factors  of  2a  +  26.  Ans.  2(a  +  6). 

2*  Find  the  factors  of  ax-\-bx-\-cx,  Ans.  (a  +  6+c)a;. 

3.  Find  the  factors  of  6a;'' +  12a; —18.  Ans.  6(a;V2ar— 3). 
4*  Find  the  factors  of  a;y+y.  Ans,  (a;+l)y. 
5*  Find  the  factors  of  aca;4-«6a;.  Ans.  (c+6)aa;. 
6i  Find  the  factors  of  3a;''y  +  3a;y'.  Ans.  3a;y(a;+y). 

7,  Find  the  factors  of  6a'4-10a'6+5a6». 

Ans.  5o(a"4-2a6  +  6'). 

8.  Find  the  factors  of  aa;"»— 6a;'^\ 

Ans.  (ax—bx^)x'^-\  or  (a— 6a;)a;*. 


THEOEEMS  AND   PACTOKINQ.  77 

9«  Find  the  factors  of  14a;y-21a?y.  Am.  1(2  — dsi)^'^', 

10.  Find  the  factors  of  51x'  — I7a;»  +  34aj. 

Ans.  11{3x^—x+2)x, 

11.  Find  the  factors  of  ea^'^'b'c'+^—6a''^bVd^\ 

Ans.  6(a''+'b'-'c^'-'d'-')a^'^b'c\ 

12.  Find  the  factors  of  2a'^6"^V— 4a'"6"'^'c'e?  +  2a"+'6'"c+6a'^* 
6"^V^  Ans.  2{a^'-2ab"'cd+a'b  +  dc^')ar-'b'^'c. 

PROBLEM 

(124.)  1.  Find  the  factors  of  2a'x— 4aV  +  2aV. 

SOLUTION. 

By  problem  B,  we  have  2a*a;— 4aV  +  2aV=2a'a;(a'— 2aV+a;'); 
but  by  Theorem  V.  (a«— 2aV +«•)=(«'—«')» ;  and  by  Theorem 
X,  a''-^x'={a-x)(a''+ax+x^);  whence,  {a'-xyz={a^xy{a'-\' 
ax  +  xy.  Therefore,  2a^x—  4a V  +  2a V = 2a'a:(a— a:)''(a'' +aa? + »»)» 
=2a''ip(a— ar)(a-a;)(a'^+aa;+a;'')(a'+aa;+a;'). 

PROBLEM 

2.  Write  the  product  of  a-\-y  +  x  by  a—y-\-x. 

SOLUTION. 

Since  a  +  y-\-xz=i{a-\-x)-\-y,  and  a— y  +  a;=(a+a;)— y,  we  have 
only  to  find,  considering  {a-\-x)  as  a  single  quantity,  the  product  of 
the  sum  and  difference  of  two  quantities.  Then  by  Theorem  m.,  we 
have  {a+y-\-x){a—y-\-x)={a-\-xy—y*. 

PROBLEM 

8.  Find  the  factors  of  [(a  +  6)'— c']-f^'. 

SOLUTION. 

Considering  what  is  within  the  brackets,  as  one  quantity,  the  given 
expression  represents  the  difference  of  two  quantities,  which  may  be 
factored  by  theorem  IX.,  X.,  or  XL  Let  us  use  Theorem  IX., 
which  gives  [{a  +  hf—c^+dWia  +  by—c^—d].  Also,  by  the  same 
Theorem,  we  have  (a+6)^— c''  =  (a+6+c)(a  +  6— c)  ;  hence,  [(a +  6)' 
— c*]"—<r'=[(a+64-c)(a  +  6-c)  +  c?][(a  +  6  +  c)(a  +  6— c)— ij. 

MISCELLANEOUS      EXAMPLES. 

1,  Square  a  +  ft+c.  Ans,  (a+6)»-f-2(a  +  6)c+c'. 


78  THEOEEMS  AND  rAOTORING. 

2,  Square  a  +  h—c—d.    Ans.  (a  +  by  - 2{a-\-b){c  +  rf)  +  (c+d)*. 
3t  Multiply  a  +  6+c  by  fe—a+c.  Ans.  (J+c)"— a*. 

4.  Factor  (a +  i)'-(c-c?)'. 

Ans.  (a  +  b—c  +  d)[(a  +  by+{a-{-b){c—d)+(c--d)'^. 

5.  Factor  m'  +  2mw  +  w'--a'+2a6— 6'. 

u4?w.  (m+w+a— 6)(m+w— a+6). 

6.  Factor  {x+t/Y—r\ 

Am.  {x+y+r^){x+y-r%{x+yy+r'}. 

7#  Factora*+(&  +  c)". 
Ans.  {a-\-b+c)[a*—a\b  +  c)  +a*(6+c)'— a(6  +  c)'  +  (6  +  c)*]. 

8i  Factor  ctc+ad  +  bd  +  bc. 

Ans,  a{c+d)  +  b{c-{-d)={a  +  b){c+d), 

9»  Factor  am  +  26a; +2<w;+6w.  ~    Ans.  (a+b){m+2x). 

10.  Factor  Sa'  +  lOa'^  +  Soft'.  -4n«.  5a{a+b)(a  +  b). 

11.  Factor  3ar'+6a^+3y'.  ud/is.  8(x+t/){x+j/). 

12.  Factor  a^—ah^.  Ans.  a{a'\-b){a-'b), 

13.  Factor  a?'— a;'y— iry^+y".  .4w«.  (a?+y)(a?— y)(a?— y). 

14.  Factor  a''-2a''6  +  2a6'-6'.  Ans.  (a— 6)(a''-a6  +  6».) 

15.  Factor  a^+a'¥  +  b\  Ans.  (a*+a6  +  6')(a''— a6  +  6'). 

16.  Factora'— 3a''a;  +  3aa;'— a?*.  Ans.  {a—x){a'-x){a—x), 

17.  Factor  Ya;''-12a;+5.  Ans.  (a?-l)(7ar-6). 

18.  Factor  «'— a;'^2aj.  Ans.  a;(a;  +  l)(a?— 2). 

19.  Factor  23?"  +  3a;»  +  x.  Ans.  a;(a;  + 1)  (2a; + 1). 

20.  Factor  a' -.6»-c''-2Jc.  Ans.  (a  +  6+c)(a-6-.c). 

(125.)  USEFUL     FORMULAS. 

1.  {a  +  by=a^-h2ab  +  b\ 

2.  {a-by=a'-^2ab-^b\ 

3.  {a-{-b)(a-b)=za'-b\ 

4.  (a'  +  a6-f-6»)(a~6)=:a«-6». 


THEOREMS  AND   FACTORING.  79 

5,  {a'-ab  +  b'){a-\-b)=a'  +  b\ 

6,  {a'-a'b-^ab'-b'){a  +  b)=a*-b\ 

7.  a'  +  2ab-\-b''={a-^b){a  +  by 

8.  a'-2ab  +  b'=(a-b){a'-b), 

10.  a'-b'={a-b){a'  +  ab  +  by 

12.  a«-6«=:(a-6)(a^  +  a'6  +  a'^>'  +  aV4-a&*  +  &')- 

13.  a'-b'  =  {a  +  b){a'-a'b-{-a'b'-a'b'  +  ab'-by 

14.  a«-6«  =  (a''-6'0K  +  «'^'  +  **)- 

15.  a'-b'={a'  +  b'){a'-¥), 

16.  a''-'6«=(a  +  6)(a-6)K  +  a&  +  6')(«'-«*  +  *'). 

17.  a*  +  a'^6''  +  6*=(a'  +  a6  +  «>')(«'-«&  +  ^')- 

18.  a'-hb':^(a-\-b)(a'-~-ab-^by 

19.  a^  +  a*6  +  a'6'^  +  a'6'  +  a6*  +  6'*=:(a  +6)(a''  +  a6  +  6'')(a'-a6  +  6') 

20.  a^-a*6  +  a'6'-a'6'  +  a^*~2»'=(«-^)  (a^'  +  aft  +  J")  (a'-aft  +  fi") 
={a-b){a'  +  a'b'  +  b'). 


CHAPTER  IV. 
GREATEST  COMMON  DIVISOB. 

(126.)  A  multiple  of  a  quantity,  is  a  quantity  that  contains  it  an 
an  exact  number  of  times.  Thus,  6  is  a  multiple  of  2  and  3,  and  ah 
is  a  multiple  of  a  and  h. 

(1  J27«)  A  measure  of  a  quantity,  is  a  quantity  that  is  contained  in 
it  an  exact  number  of  times.  Thus,  2  and  3  are  measures  of  6,  and  a 
and  h  are.  measures  of  a6. 

(128.)  Kcommmi  measure  or  common  divisor  of  two  or  more 
quantities  is  one  that  is  exactly  contained  in  each  of  them. 

(129.)  The  greatest  common  measure  or  greatest  common  divisor 
of  two  or  more  quantities  is  the  greatest  quantity  that  is  exactly  con- 
tained in  each  of  them. 

THEOREM     I. 

(130.)  If  a  quantity  measures  another  quantity  it  will  measure 
any  multiple  of  that  quantity, 

DEMONSTRATION. 

Let  a  be  a  quantity,  then  ta  is  a  multiple  of  a. 

We  are  to  prove  that  if  a  quantity  d  measures  a  it  will  also 
measure  ta.  Let  r  be  the  number  of  times  d  is  contained  in  a; 
whence,  we  have,  a—rd  and  ta^ztrd.  Now  d  measures  trd  and  must, 
therefore,  measure  to,  which  is  equal  to  trd. 

THEOREM     II. 

(131.)  If  a.  quantity  measures  two  other  quantities,  it  will  also 
measure  their  sum  and  their  difference. 

DEMONSTRATION. 
Let  Rd  and  rd  be  two  quantities  that  are  divisible  by  d. 


GKEATEST   COMMON   DIVISOR.  gl 

We  are  to  prove  that  Bd  +  rd  and  Rd—rd  are  both  divisible  by  d» 
Since  Bd+rd=d{R-\-r)  and  Rd—rd=d{R—r\  we  see  that  d  is 
a  factor  in  both  these  expressions,  and  is,  therefore,  a  divisor  of  both. 

PROBLEM. 

(132.)  To  find  the  greatest  common  divisor  of  two  or  more 
monomials. 

RULE. 

Resolve  the  monomials  into  their  factors,  and  the  product  of  the 
fojctors  common  to  all  the  mcmomials  will  he  the  greatest  common 
divisor. 

PROBLEM. 

(133.)  Find  the  greatest  common  divisor  of  3a^b^d\  6ab^d\ 
and  l^bd', 

^    8 OLUTI O  N. 

Resolving  these  monomials  into  factors,  we  have 

3a'b'd*=  a'bd''Sbd\ 

6ab'd'=2abd  '  Bbd% 

and    12bd^=        4  •  36c?'. 

From  which  it  is  seen  that  Sbd^  contains  all  the  common  factors, 

and  is,  therefore,  the  greatest  common  divisor. 

EXAMPLES. 

1,  Find  the  greatest  common  divisor  of  12a6'c  and  256V. 

Ans.  b*e, 

2,  Find  the  greatest  common  divisor  of  3a'6V  and  6a*6V'. 

Ans.  3a'6V. 

3*  Find  the  greatest  common  divisor  of  3a^b~^  and  2abcd^, 

Ans.    abh 

4.  Find  the  greatest  common  divisor  of  49a^h'^c*  and  63a'6'c'. 

Ans.     la'bY. 

5*  Find  the  greatest  common  divisor  of  ai  and  ai. 

6«  Find  the  greatest  common  divisor  of  a-i  and  ah 

7,  Find  the  greatest  common  divisor  of  a-i  and  -i. 

Remark. — The  last  three  examples  are  left  unanswered  to  call  out  thought. 


82  GEEATEST  COMMON  DIVISOR. 

PROBLEM. 
(134.)  To  find  the  greatest  common  divisor  of  two  polynomials. 

RULE. 

1.  Find  the  greatest  monomial  factor  that  is  contained  in  both  poly- 
nomials,  and  reserve  it. 

2.  Meject  the  remaining  monomial  factors  from  each  polynomial. 

3.  Arrange  the  terms  of  the  resulting  polynomials  according  to  the 
powers  of  some  letter  in  both,  and  consider  that  polynomial  of  which 
the  leading  letter  in  the  first  term  has  the  least  exponent  as  the  divisor, 
and  the  other  polynomial  as  the  dividend. 

4.  Multiply  the  dividend  by  the  lea^t  monomial  that  will  render 
the  first  term  of  the  dividend  exactly  divisible  by  the  first  term  of  the 
divisor, 

6.  Divide  the  dividend  by  the  divisor^  and  continue  the  division 
until  the  highest  exponent  of  the  leading  lefter  in  the  remainder  is  less 
than  the  highest  exponent  of  the  leading  letter  in  the  divisor.  [^If  the 
coefficient  of  the  first  term  in  any  remainder  is  not  divisible  by  the  co- 
•  efficient  of  the  first  term  in  the  divisor,  to  avoid  fractions  multiply  the 
remainder  by  such  a  number  a^  will  render  its  first  coefficient  exactly 
divisible  by  the  coefficient  of  the  first  term  of  the  divisor^ 

6.  Reject  from  the  remainder  its  greatest  monomial  factor,  and  then 
consider  the  result  a  new  divisor,  and  the  former  divisor  a  new  divi- 
dend, proceed  cw  before,  and  continue  the  process  until  the  remainder 
is  zero. 

7.  Multiply  the  last  divisor  by  the  reserved  monomial,  and  the  pro- 
duct will  be  the  greatest  common  divisor  of  the  given  polynomials. 

DEMONSTRATION. 

Let  A  and  B  represent  two  polynomials  of  which  we  seek  the  greatest 
common  divisor.  Let  C  and  D  represent  two  other  polynomials, 
neither  of  which  can  be  divided  by  a  monomial. 

Suppose  A=a^bC  audi  B^zo^cD  ;  a,  b,  and  c  being  monomials. 

1.  It  is  evident  that  the  greatest  monomial  factor  common  to  a^bO 
and  a^cD  is  a^,  which  must  be  reserved,  because  it  evidently  is  a  factor 
of  the  greatest  common  divisor  of  A  and  B,  or,  which  is  the  same, 
of  «'6(7and  aVD. 


GREATEST   COMMON   DIVISOR.  83 

2.  We  have  left  ahC  and  cD.  We  now  seek  the  greatest  common 
divisor  of  these  polynomials.  Since  a  is  not  a  factor  common  to  both 
these  polynomials,  it  can  not  be*  a  factor  of  their  common  divisor,  and 
therefore  may  be  rejected.  For  the  same  reason  h  and  c  may  be  re- 
jected. Hence  the  greatest  common  divisor  of  ahC  and  cD  is  the 
same  as  the  greatest  common  divisor  of  C  and  D. 

3.  Suppose  that  the  terms  of  the  polynomials  0  and  D  are  arranged 
according  to  the  powers  of  the  same  letter  in  both,  and  that  the  ex- 
ponent of  the  leading  letter  in  the  first  term  of  C  is  less  than  the  ex- 
ponent of  the  same  letter  in  the  first  term  of  D.  Therefore,  consider 
(7  as  a  divisor  and  i>  as  a  dividend. 

4  Suppose  the  first  term  of  C  contains  the  monomial  3m  and  that 
the  first  term  of  D  does  not.  Then,  in  order  that  the  first  term  of  the 
quotient  should  not  be  fractional,  the  polynomial  D  should  be  multi- 
pHed  by  3?7i.  The  greatest  common  divisor  of  C  and  D  is  the  same* 
as  the  greatest  common  divisor  of  C  and  3mi>,  since  the  introduced 
monomial  3m  can  form  no  part  of  the  greatest  common  divisor  of  C 
and  3mZ),  because  by  hypothesis  3m  can  not  be  a  factor  of  C. 

Operation.  5.  Divide  Bml)  by  C,  and  suppose  the  first  ferm 

C)3mD{q  of  the  quotient  to  be  q,  and  the  first  remainder 

qC  E.    Again,  suppose  that  the  first  coefficient  of 

E  C  contains  the  factor  2,  and  that  the  first  co- 

2  eflScient  of  E  does  not.     Then  multiply  E  by  2 

C)2E{q'  and  divide  the  result  2E  by  C  and  let  q'  repre- 

q'O  sent  the  first  term  of  the  quotient,  and  i^  the 

bp)  F  remainder.     Also  suppose  that  the  exponent  of 

G)2nC{q"  the  leading  letter  in  the  first  term  of  i^is  less 

q"G  than  the  exponent  of  the  same  letter  in  the  first 

0  term  of  C. 

6.  Suppose  that  the  greatest  monomial  factor  of  ^is  bp.  Reject 
this  factor,  or,  in  other  words,  divide  F  by  5p,  and  let  G  represent 
the  result,  which  consider  as  a  new  divisor  and  C  as  a  new  dividend. 
Suppose,  then,  the  first  term  of  G  contains  the  monomial  2n  and  the 
first  term  of  C  does  not.  Then,  in  order  that  the  first  term  of  the 
quotient  should  not  be  jfractional,  the  polynomial  O  should  be  multi- 
phed  by  2n. 

Let  q"  be  the  exact  quotient  of  2nC  by  G. 

7.  6^^  is  the  greatest  common  di/isor  of  C  and  i>,  and  multiplying 


84  GREATEST  COMMON   DIVISOR, 

it  by  the  reserved  monomial  «",  we  have  a^Q  for  the  greatest  com- 
mon divisor  of  A  and  B, 

We  prove  that  G  is  the  greatest  common  divisor  of  C  and  D  as 
follows : — 

Let  G'  be  the  greatest  common  divisor  of  C  and  B.  We  have  al- 
ready shown  that  the  greatest  common  divisor  of  C  and  ^nD  is  the 
same  as  the  greatest  common  divisor  of  C  and  D. 

Since  G'  is  a  measure  of  (7  and  D,  it  must  (130)  be  a  measure 
oi  qC  and  3mi>,  and  it  must  also  ( 1 3 1 )  be  a  measure  of  JS,  the  dif- 
ference between  qC  and  3mi). 

Since  G'  is  a  measure  of  C  and  ^,  it  must  also(130)  be  a 
measure  of  gf'Cand  2E^  and  it  must  also  (131)  be  a  measure  of  i^, 
the  difference  between  q'  C  and  2E. 

Because,  by  hypothesis,  C  and  D  contain  no  monomial  factors,  it 
follows  that  G',  their  greatest  common  divisor,  is  neither  a  monomial 
nor  divisible  by  a  monomial. 

But  G'  measures  F,  and  consequently  must  measure  G,  which  re- 
presents F  after  its  monomial  factors  are  rejected.  Hence,  the  great- 
est common  measure  of  C  and  D  can  not  be  greater  than  G,  and, 
therefore,  if  (r  is  a  common  measure  of  0  and  i>,  it  must  be  the 


Since  G  measures  2nO^  and  is  not  a  monomial,  it  must  also  mea- 
sure (7,  and  consequently  must  measure  5'' (7. 

But  G  also  measures  F^  therefore  it  must  measure  2E,  which  is  the 
sum  of  F  and  q'C.  Because,  G  measures  2-fi',  and  is  not  a  monomial, 
it  must  also  measure  F :  G  must  also  measure  qC^  and  therefore  must 
measure  3wiZ>,  which  is  the  sum  of  E  and  qC.  Since  G  measures 
3mZ>,  and  is  not  a  monomial,  it  must  also  measure  D,  Hence,  G  is 
a  common  measure  of  C  and  i>,  and  must,  therefore,  as  shown  above, 
be  the  greatest. 

Now  let  C—QG2indiD=  Q'G,  and  we  have 

A=a^bQG  and  JS=a^cQ'G ;  whence,  we  see  that  a^G  must  be 
the  common  measure  of  A  and  B,  because  Q  and  Q'  may  be  rejected, 
since  they  can  have  no  common  measure. 

Therefore  the  truth  of  the  rule  is  established. 


PROBLEM. 

(135»)  Find  the  greatest  common  divisor  of  the  polynomials, 
6a'b~6a'bi/—2by' -{-2aby''{A)  and  lia'b  +  Sbf-ldabt/  (B). 


€^REATEST  COMMON  DIVISOR.  85 

SOLUTION. 

I*  We  see  by  inspection  that  b  is  the  greatest  monomial  that  is 
common  to  both  polynomials,  which  reserve. 

2.  By  rejecting  the  monomial  2  from  (^),  and  3  from  (J5),  we 
have  3a^—Sa^j/—y^  +  ay^  and  4a''  +  2/'* — hay. 

3*  Arranging  these  results  according  to  the  powers  of  the  letter  a, 
we  have  3a^ — Za^y  +  ay"^ — y^  and  4a'' — hay + y'.  Hence,  the  latter 
must  be  the  divisor. 

Operation, 

3a'-  3aV+     oy'~     f 
4 


4a'— 6ay+y»)12a'— 12aV+  4ay*—  4/(8a 

12a''— 15a  V+  ^ay"" 

Za^y^     ay'-  4/ 
4 


12aV+   4ay^-16yX8y 
12aV— 15ay''+   3/ 


19y')19ay^l92^ 

a— y)4a'— 6ay+y*(4a— y 
4a'— 4ay 

—  ay  +  y' 

0 

4(  Since  the  first  term  of  the  divisor  is  not  contained  in  the  divi- 
dend, we  multiply  the  dividend  by  4,  and  then  dividing,  we  have  for 
the  first  remainder  Sa'zz  +  ay''— 4?/^,  which  must  be  multiplied  by  4  to 
avoid  a  fractional  result  in  the  quotient.  Continuing  the  division,  we 
have  for  the  next  remainder  19ay'  — 19y',  in  which  the  exponent  of 
the  leading  letter  a  is  less  than  the  exponent  of  the  same  letter  in  the 
first  term  of  the  divisor. 

5t  Rejecting  from  this  remainder  IQy'*,  the  greatest  monomial  con- 
tained in  it,  we  have  a— y,  which  constitutes  the  new  divisor,  which 
divided  into  4a'— 5ay  +  2/^,  leaves  zero  for  a  remainder. 

7t  Hence,  the  last  divisor  a—y  multiplied  by  the  reserved  mono- 
mial 6,  gives  h(a—y)  for  the  greatest  common  divisor  of  the  given 
polynomials  (A)  and  (B). 


,86  GREATEST   COMMON   DIVISOB. 

EXAMPLES. 

1.  Find  the  greatest  common  divisor  of  ic*— 8a;' +  2 la;'— 20a; +  4 
and  23;"— 12a;' +  2  la;— 10.  Ans.  x—2, 

2.  Find   the   greatest   common   divisor  of  Sa'*— 2a— 1    and   4a" 

— 2a'^— 3a  +  l.  Ans.  a—1. 

« 
3«  Find   the  greafest   common  divisor   of    a^—b^   and  a^  +  2a'6 

H-2a6'  +  6'.  Ans.  a''  +  ab  +  b\ 

4.  Find  the  greatest  common  divisor  of  a*-\-a^b—ab^—b*  and 
a^+d'b'  +  b\  Ans.  a''  +  ab  +  b\ 

5t  Find  the  greatest  common  divisor  of  a''  — 2aa;  +  a;''  and  a^—a^x 
— ax^+x^.  Ans.  a'  — 2aa;  +  a;'. 

6«  Find  the  greatest  common  divisor  of  a;'  +  9a;^  +  27a^— 98  and 
a;' +  12a;— 28.  Ans.  x—2. 

7.  Find  the  greatest  common  measure  of  Va'— 23a6  +  66'  and  5a* 

—  18a''b  +  llab^—6b\  Ans.   a—Sb. 

8.  Find    the    greatest    common    divisor    of    6a®  +  15a*6— 4a''c' 

—  lOa'bc''  and  9a'6— 2'7a'6c— 6a6c''  +  186c'.  Ans.  3a»— 2c'. 

9.  Find  the  greatest  common  divisor  of  SGa'^'- 18a'6'— 27a*6' 
+  9a'6' and27a'6'— 18a'6'-9a'6'.  Ans.  9a%\a—l). 

10.  Find   the  greatest  common   divisor  of  {c—d)a^  +  {2bc—2bd)a 
-{-(b'c-b'd)  and  {bc-bd  +  c'-cd)a  +  (b'd  +  bc'-b'c-bcd). 

Ans.  c—d. 

PROBLEM. 
(136.)  To  find  the  greatest  common  divisor  of  two   or  more 
polynomials. 

RULE. 

Resolve  all  the  polynomials  into  their  simplest  factors^  and  take 
the  product  of  those  which  are  common,  for  Hie  greatest  common 
divisor. 

DEMONSTRATION. 

The  truth  of  this  rule  is  a  consequence  of  the  definition  of  the 
greatest  common  divisor. 

PROBLEM. 

(137.)  Find  the  greatest  common  divisor  of  a;'— 6a;  +  2aa;— 2a6 
and  x^  4-  ax^  +  bx^ — 2a^x  +  bax — 26a'. 


.      GEEATEST   COMMON  DIVISOR.  87 

SOLUTION. 

Arranging  x'^—hx  +  2ax—2ah^  we  have  x"^ -\-2ax—hx~2ah. 

Factoring,  we  obtain  {x-\-2a)x—h(x  +  2a)  =  (a;  +  2a){x—h), 

Arranging,  x^ -\-ax'^  +  hx'^—2a^x-\-  hax—2ha^,  we  have  x^-{-ax^ 
— 2aV  +  hx""  +  bax—2ba^. 

Since,  ax^=2ax^ — ax^  and  bax=:2hax — bax^  we  have  x^-\-ax^  —  2a'x 
+  bx'^  +  bax—2ba^=x^  +  2ax^—ax'^  —  2a^x  +  bx^  +  2bax—bax—2ba^. 

Factoring,  we  obtain  {x  +  2a)  x^—ax{x-\-2a)  +  bx  (x  +  2a)  —  ab 
(x  +  2a)  =  (x  +  2a)  (x"^ — ax  +  bx— ab) . 

But,  x'^—ax  +  bx—ab=(x—a)x-\-b{x—a)  =  (x—a)(x-\-b). 

Hence,  we  have  a;'H-aa;''  +  6a;''— 2a''a;  +  6aa;— 25a''=(a;  +  2a)(a;— a) 
(x  +  b). 

We  also  have  x^—bx  +  2ax—2ab=(x-\-2a){x—b). 

Whence  we  see  that  a;  +  2a  is  the  only  common  factor. 

EXAMPLES. 

1.  Find  the  greatest  common  divisor  of  a:'— a",  x*—a*,  and  x^^a^, 

Ans.  x—a. 

2.  Find  the  greatest  common  divisor  of  a*— a;*,  a'— a^r— aa;^  +  aj', 
and  2a^b—2abx^.  Ans.  a^—x^. 

3.  Find  the  greatest  common  divisor  of  5a^  +  10a*b  +  5a''b^j  a'S^- 
2a'b''  +  2a6'  +  b\  and  a^—b\  Ans,  a  +  b. 

4.  Find  the  greatest  common  divisor  of  x^  +  ax^^a^x—a*  and 
a;*4-aV  +  a*.  Ans.  x''  +  ax  +  a\ 

5.  Find  the  greatest  common  divisor  of  ar*— 1,  x^+x^^  a;*— 1,  and 
a:*  +  2a;'  +  l.  Ans.  x^  +  1. 

6.  Find  the  greatest  common  divisor  of  x^—b^x  and  x^-j-2bx-\-b^. 

Ans.  x  +  b. 

7.  Find  the  greatest  common  divisor  of  Sa'—3a'6  +  a6'— 6' and 

4a'b—5ab^  +  b^.  Ans.  a—b. 

8.  Find  the  greatest  common  divisor  of  a'  — 5a6  +  46',  a^—d'b 
-\-3ab'^  —  Sb^,aiida*—b*.  Ans.  a—b. 

9.  Find  the  greatest  common  divisor  of  12a;*— 24a:^y  +  12a;y  and 
8ajy— 24a;y4-24a'y*— 82/'.  Ans.  4:(x^—2x7/  +  f). 

10.  Find  the  greatest  common  divisor  of  2a;'' +  3aa;  +  a",  2aa;'*— a'a; 
—a",  and  6a;'^4-3aar.  Ans.  2x+a. 


CHAPTER    V. 
LEAST  COMMON  MULTIPLE. 

(138.)  The  LEAST  COMMON  MULTIPLE  of  two  OT  more  quantities 
is  the  least  quantity  that  can  be  measured  by  all  of  them. 

PROBLEM. 

(139.)  To  find  the  least  common  multiple  of  two  or  more 
quantities. 

RULE. 

Resolve  the  given  quantities  into  their  simplest  factors,  and  find  the 
continued  product  of  all  the  different  factors,  taking  each  factor  the 
greatest  number  of  times  that  it  occurs  as  a  factor  in  any  of  the  given 
quantities,  and  the  result  will  he  the  least  common  multiple  sought. 

DEMONSTRATION. 

The  least  common  multiple  must  contain  as  many  factors  as  are 
contained  in  any  of  the  given  quantities. 

PEOBLEM. 

(140.)  Find  the  least  conmaon  multiple  of  a'— a;',  a'— 2aaJ+ar*, 
and  a'^^x^. 

SOLUTION. 

Factoring  these  quantities,  we  have 
{a-\-x)(a^x),  {a—x){a—x),  and  (a — x){a  +  x){a^ -\- x^). 

Hence,  the  least  common  multiple  is  (a  +  a;)(a— a:)(a— ar)(a'  +  a;') 
= a' — ax* — a*x  +  «*. 

EXAMPLES. 

1.  Find  the  least  common  multiple  of  2a'ar  and  lOaV. 

Ans.  lOa'a;'. 

2i  Find  the  least  common  multiple  of  Sax',  4:a^h^x,  and  ^a^h^xy. 

Ans.  1Aa^h^x*y. 


LEAST  COMMON   MULTIPLE.  89 

3i  Find  the  least  common  multiple  of  x^—y"*  and  «'— y'. 

Ans.  (a;+y)(a;'— y"). 

4*  Find  the  least  common  multiple  oia^—h^  and  a^  +  &^ 

Am.  (a-6)(a«+6»). 

5*  Find  the  least  common  multiple  of  a;'— 8a; +  '7  and  x^  +  ^x—Q, 

Ans.  ic^— 5*70: +  56. 

6«  Find  the  least  common  multiple  of  a;'— aj'y—a^y'  +  y",  x^—x^y  + 
xy^—y^  and  a;*— y*.  ^W5.  x^—xy*—x*y+y^. 

7.  Find  the  least  common  multiple  of  (a +  6)',  a^—b^^  {a— by,  and 

8.  Find  the  least  common  multiple  of  a +6,  a— 6,  d'  +  ct6  +  6',  and 
a'^ab  +  b'.  Ans.  a'—b\ 

9i  Find  the  least  common  multiple  of  a^+^a'b  +  ^ab^  +  b^,  a'  + 
2a6  +  6'',  and  a'— 6'.  Ans.  a*  +  2a'&— 2a6'— 6*. 

10.  Find  the  least  common  multiple  of  a;*— Sa-'+Qa;"— '7a;  +  2,  and 
x^—Qx'  +  Sx—Z.  Ans.  a;*— 2a;*— 6a:'  +  20a;''— 19a;+6. 


CHAPTER  VI. 
ALGEBRAIC  FRACTIONS. 

(141.)  An  Algebraic  Fraction  is  an  algebraic  expression  de- 
noting the  di\dsion  of  one  quantity  by  another,  and  is  written  in  the 
same  manner  as  a  common  fraction  in  arithmetic. 

An  algebraic  fraction  may  also  be  considered  as  a  certain  part  of 
unity. 

THEOREM    I. 

(14:3«)  The  value  of  a  fraction  is  not  changed  when  both  of 
its  terms  are  multipHed  by  the  same  quantity. 

DEMONSTRATION. 

A 
Let  -^  represent  any  algebraic  fraction. 

We  are  to  prove  that  -=r-= — =r. 
^  B     mB 

By  (101),  we  have  — g=  — ^ — ~'~B~~~W^  because  m°=l. 

THEOREM    II. 

(143.)  The  value  of  a  fraction  is  not  changed  when  both  of  its 
terms  are  divided  by  the  same  quantity. 

DEMONSTRATION. 
Let  -5-  represent  any  algebraic  fraction. 

We  are  to  prove  that  — -=— -^ — . 
B     B-\-m 

Since,  A-\-m= — and  ^-T-m= — ^^  we  have,  by  (101)  — r7»= 

mr^m^A    m^A    A       ^   „  ^ 


ALGEBRAIC  FRACTIONS.  91 

PROBLEM. 
(144.)  To  reduce  a  fraction  to  its  lowest  terms. 

RULE. 

Resolve  both  terms  into  their  simplest  factors,  and  cancel  those  that 
are  common. 

Or,  divide  both  terms  by  their  greatest  common  divisor, 

DEMONSTRATION. 
The  accuracy  of  this  rule  depends  upon  (1 43). 

PROBLEM. 

(145.)  Reduce  —. ^ 5 x  to  its  lowest  terms. 

^  ^  x^+ax^—a^x—a* 

SOLUTION. 

Factoring,  we  have 

a;*  +  aV  4- a*         _{x^  +  a^  +  ax){x^-{-a^—ax)__ 
x^-\-ax^—a^x—a^  ~  x^—a^  +  ax^—a^x  ~~ 

(x^-\-<f+ax)(x^+a^—ax)       ^(x'*-{-a^  +  ax)(x'  +  a^—ax) 
(x^  +  a''){x''-a'')+ax(x^-a^)~'(x^-{-a^+ax)lx^-a^)         ' 

Canceling  the  common  factor,  there  results 


x'—a'  X'- 

EXAMPLES. 


21 

1.  Reduce  — rr^  to  its  lowest  terms.  Atis.  — r. 

5a^b^  5b 

2.  Reduce -„  to  its  lowest  terms.  Ans. . 

ax+x  a+x 

„    ^  ^        14a;y— 21rcy       .     ,         ,  ,        2y— 3rcy 

d.  Reduce —-z to  its  lowest  terms.      Ans.  -^ -, 

IX  y  X 

4.  Reduce    ,  .  ^-^  to  its  lowest  terms.  Ans. 


x^-a^x^ 


5*  Reduce  — — -^j^  to  its  lowest  terms.  Ans.  — j — . 


92  ALGEBRAIC  FRACTIONS. 

2a;' 16a; 6  2 

6«  Reduce  —-5 — r- to  its  lowest  terms.  Ans,  -. 

7t  Reduce  — ^ ^— — ^  to  its  lowest  tenns.       Ans, 


St  Reduce    .  .  ^  , — rr-  to  its  lowest  terms.  Ans.  -.-. 

a^+2ab-{-b^  a  +  b 

a+26  +  36' 


9.  Reduce  —7 — — ^^ ttt-  to  its  lowest  terms. 

2a*— 3a  0— 5a  0 


Ans, 


2a''— 3a6— 66'* 


a'— 2a'6  +  2a6'— 6^        .     , 

10.  Reduce 1 ^tttti to  its  lowest  terms. 

a*  +  a*6^+6* 

a—b 
Ans, 


a'+ab  +  b^ 
y* /J.*  y^-\-X^ 

11  •  Reduce  -5 5 ^—5- — 7  to  its  lowest  terms.      Ans,  . 

y^—y^x—yx-\-x^  y—x 

/p2^2a;— 3  a?— 1 

12t  Reduce  -^ — to  its  lowest  terms.  Ans,  — -— . 

x^-\-5x-\-Q  x  +  2 

to    T>  J         2a;'  +  3a;^  +  a;^    v   i         ..  a  ^^  +  1 

13*  Reduce  — = 5 — -— to  its  lowest  terms.  Ans,    — -— . 

x^—x^—2x  x  +  2 

%M    -n  n       ac-\-bd+ad-\-bc      ^    .^   ,         ,  . 
II.  Reduce  -^ .  ^,      - — r-r?  to  its  lowest  terms. 
a/^+26a;4-2aa;-|-6/^ 

15.  Reduce — -. — 5 — -7 to  its  lowest  terms. 

a^—b^—c^—2bc, 

.       a-hb+e 

Ans.  = — . 

a — 6 — c 

^o    T.  J        a'-3a&4-ac  +  26'-26c  ^    .^   , 

16.  Reduce  5 — ,„  .  ^, 5 to  its  lowest  terms. 

a  — 0  +26c— c 

a— 26 

Ans,  r — . 

a  +  6— c 

—    ^   ,  (a  +  6)(a  +  6+c)(a  +  6— c)         ,    .,   , 

"•  ^^^^  2a'6-  +  2aV  +  26V^a^3Fi:?-^"  ^^  ^^^^^'  *^"^- 

Ans, 


(c  +  a— 6)(6— o  +  c)* 


ALGEBRAIC  FBACTTOKS.  93 


18.  Reduce  — — 77-^ — ( =-  to  its  lowest  terms.         Ans, 

x^-^(b  +  c)x  +  bc 


*A    1.  J         2x'-\-x'-Sx  +  5  ^    .. 

19.  Reduce  t-^ — — —  to  its  lowest  terms. 

^x^  —  l2x-\-5 

Ans.  — — . 

PROBLEM. 

(146.)  To  reduce  a  fraction  to  an  entire  or  mixed  quantity. 

RULE. 

Divide  the  numerator  by  the  denominator,  and  when  there  is  a  re- 
mainder  place  it  over  the  denominator  for  the  fractional  part,  and 
connect  it  with  the  entire  part  of  the  quotient  by  the  sign  of  addition, 
or  change  all  the  signs  of  the  numerator  of  the  fractional  part,  and 
connect  it  with  the  entire  part  of  the  quotient  by  the  sign  of  sub- 
traction, 

DEMONSTRATION. 

The  accuracy  of  this  rule  is  an  obvious  consequence  of  the  accuracy 
of  the  process  of  division.  » 

PROBLEM. 

(1 47 .)  Reduce  ^y  — y  +a;y— y— a;+     ^  ^  ^^^ quantity. 

SOLtTTION. 

Dividing  the  numerator  by  «'— 1,  we  get  y'  and  the  fractional 
,  xy'-y-x-{-\_{x-\)y-l{x-l)_{x-\)(y-^l)_y--\ 

connected  with  y'   by   the   sign   of  addition  gives    y^  +  ~~Tj  ^^'^ 

X  -x"  X. 

with  the  sign  of  subtraction  gives  y' -,  or  y^—z — -, 

X -\-  A  1  -x'X 

EXAMPLES. 

1,  Reduce  — ^ —  to  a  mixed  quantity.  Ans,  a+j. 

2*  Reduce to  a  mixed  quantity.  Ans.  aH , 

a  ^         J  a 


94'  ALGEBRAIC   FRACTIONS. 

3*  Reduce to  a  mixed  quantity.     Ans.  5a  +  l-\ — — . 

,    ^   ,        4a  +  2ax  +  b  .     ,  .  j        .     ^       ^ 

4*  Reduce to  a  mixed  quantity.      Ans.  44-2a;H — . 

.,    -.^   ,       x'y—xy—x+1  .     ,  .  ^  1 

o*  Reduce  —^ to  a  mixed  quantity.         Ans,  x , 

xy—y  ^  y 

6.  Reduce  •— to  a  mixed  quantity.         Ans.  1— a — . 

\^x-^*J  a;-4-2 

7.  Reduce  — - —  to  a  mixed  quantity.      Ans,  Zx-V\-\ — — -. 


8.  Reduce to  a  mixed  quantity. 

ox 


Ans.  ^x -. 

6x 


T9.  Reduce to  a  mixed  quantity. 

oCI/ 

Ans.  a— 3c— 2H , 

3a 

10.  Reduce  to  an  entire  quantity.   Ans.  5{x^-{-xy-{-y'*), 

X    y 

PROBLEM. 

(1 48.)  To  reduce  a  mixed  quantity  to  a  fractional  form. 

RULE. 

Multiply  the  entire  part  by  the  denominator  of  the  fractional  paA 
and  add  the  numerator  to  the  product  when  the  sign  of  the  fraction  is 
plus,  and  subtract  it  from  it  when  the  sign  of  the  fraction  is  minus  ; 
this  result  placed  over  the  denominator  will  give  the  required  form, 

DEMONSTRATION. 

This  rule  indicates  a  process  whicli  is  just  the  reverse  of  that  in 
the  last  rule,  and  must,  therefore,  be  accurate,  because  the  problem  to 
be  solved  is  just  the  reverse. 

PROBLEM. 

(149.)  Reduce  2a to  a  fractional  form. 


ALGEBRAIC  FRACTIONS.  95 

SOLUTION. 

Multiplying  2a  by  c  gives  2ac,  and  from  this  must  be  subtracted 

Bx—bj  because  the  sign  of  the  fraction  is  minus,  whence  we  have 

2a—{3x-b)     2a-Sx  +  b  .     ^,  ... 

^^ -= for  the  required  form. 

c  c 

Scholium. — It  should  be  remembered  that  the  3a;  in  this  example 
is  plus,  and  that  the  minus  sign  before  the  fraction  belongs  to  the 
whole  fraction,  or  what  is  the  same  thing,  belongs  to  the  whole 
numerator  as  indicated  in  the  solution  by  the  parentheses. 

EXAMPLES. 

d^ /^a  2a^ 

!•  Reduce  1 H — = r  to  a  fractional  form.  Ans.  -= r. 

a  +a;  a'+aj' 

x  xix  -4-  2/1 

2.  Reduce  x-\ to  a  fractional  form.  Arts,  — ^. 

y  y 

n    T.   T        .       2a;— 5  -     ..      ,  ^  .       6(2a;  +  l) 

3,  Reduce  4a; —  to  a  fractional  form.  Atis,    ^     — '-. 

3  3 

4i  Reduce  3a;— 9 ; —  to  a  fractional  form.       Ans. 


a;+3  a;  +  3 


5.  Reduce  1 H 7 to  a  fractional  form. 

26c 


26c 


6.  Reduce  1— ^ — tk-  to  a  fractional  form.  Ans, 


lab 


a^-Vb''  a" +  6" 

7.  Reduce  1 z to  a  fractional  form. 

25c 

Ans   («  +  5-c)(g-6  +  c) 
2bc 

a?  +  a;'  2a;* 

8,  Reduce  a—x to  a  fractional  form.         Ans, 


a  +  a;  a+x 

9.  Reduce  2a— a;  H-^^ —  to  a  fractional  form.  Ans,  — . 

X  X 

x^ 

10,  Reduce  a^—ax-^x^ to  a  fractional  form. 

a+x 

a' 
Ans. 


a-\-x 


96  ALGEBRAIC    FRACTIONS. 

PROBLEM. 

(150.)  To  reduce  fractions  to  equivalent  fractions  having  a 
common  denominator. 

RULE 

1.  Multiply  both  terms  of  each  fraction  hy  the  prodiict  of  all  the  de- 
nominators except  its  own* 

RULE 

2.  Find  a  commmi  multiple  (generally  the  least)  of  all  the  denominr 
ators,  and  divide  it  by  each  denominator,  and  then  multiply  both 
terms  of  each  fraction  by  these  results  resptectively, 

DEMONSTRATION. 

The  accuracy  of  these  rules  depends  upon  the  fact  that  the  value  of 
a  fraction  is  not  changed  when  both  of  its  terms  are  multiplied  by  the 
same  quantity. 

PROBLEM. 

( 1 5 1  •)  Reduce ,    -3,  and ^  to  equivalent  fractions 

having  a  common  denominator. 

SOLUTION. 

Applying  Rule  1,  we  have 

(1+^)(1_^^)(1_^3)    (1^^.)(1_^)(1_^S)    _^(l+^3)(l_^)(i_^.) 

(1  -x){\  -x'^){\  -x^y  (1  -x'){i  -x){i  -xj  ^'"''{i  -x')(i  -x){i  -xy 

Applying  Rule  2,  using  the  least  common  multiple  of  the  denom- 
inators, which  is,  (1  +x)(l-{-x^),  or  (1 -\- x){l  +  x){l  -{-x  +  x^),  we  have 
(l+x)(l+x){l+x  +  x')     {i-x'){i+x-x')  {l+x')(l+x) 

(^l^x){l+x)(l+x-\-xj  {i-x'){i+x  +  xj  ^""^  (l-rc^)(l4-ajy'''* 
by  putting  the  denominators  in  the  same  form,  we  obtain 
(^i^xY{l-{-x  +  x')  {l+x'){l+x  +  x')  (l+a;)(l+a;^) 

(l+a:)(l-a:^)    '     {l+x)(l-x')     '  ^'"'^  (l+a:)(l-a:«y 
EX  amples. 
a         c 

1,  Reduce  ~  and  -5  to  equivalent  fractions  having  a  common  de- 

.     ,  .       ad      .  be 

nommator.  Ans.  j-^  and  -=-=. 

oa         bd 


ALGEBRAIC  FRACTIONS.  97 


2«  Reduce and  — ^—  to  a  common  denominator. 


x+y         x—y 


x'—xy         xy-\ry^ 
Am,  -- — \  and  -V"^- 


3«  Reduce r, -,  and  -^ — tt,  to  a  common  denominator. 

it  Reduce , ,  and  —  to  a  common  denominator. 

a+x'     3    '         2x 

18a;'       2a'x—2x'       ,    3a  +  3a; 

j\fi8   ^——^—^——~  and.  — — ^^— -^— 

6aar  +  6a:"  6aa;  +  6a;"  6aa;-f6a;'* 

^    -,   ,        a  2c'        ,        3a— a;'  ,  .     ^ 

da  Reduce  t,  -r»  and  x-\ to  a  common  denommator. 

b   a  x 

adx  2bc'x  Sabd 

bdx^   bdx  '  hdx  ' 

X    X  -I-  \  X        \ 

6*  Reduce  -,  — : — ,  and to  a  common  denominator. 

3      4  \-\-x 

4a;'  +  4a;    3a;'  +  6ar4-3        ^  12a;— 12 

Ans,  , ,  and . 

12a;+12*     12a;  +  12    '         12a;  +  12 

7.  Reduce and    ^  ^    to  a  common  denominator. 

x  3ao 

6abx-\-9ab       ,  5a;' -{-2a! 

Ans.  — —-Z and  — — -; — . 

3a6ar  3aoa; 

8.  Reduce r, — ,  and to  a  common  denominator. 

a^—x^  4a—  4a;  a  -fa; 

4a        3a6-f  36a;  20aa;— 20a;' 

•^''''  4a'-4a;''  4^^=ix^'  ^""^  ~4a'-4a;'  ' 

-^    T>   ,         ^         ^'—5       j^     4a— 15^ 

9«  Reduce  — ,  a ,  and  7  H —  to  a  common  denommator. 

5y  Sx  2 

6x^    SOaxy—lOx'y -]-  50y       .  QOaxy—lbxy 

Ans,  -— — , — ,  and . 

30a;y  30a;y  '  30a;y 

1  1  6y 

10*  Reduce  -= z,  -5 =,  and  -i rto  a  common  denomnator. 

a:  +  2ax-\-  a;'  a' — a;'         a* — x* 

a^-[-ax'^—a'^x~x^  a^  +  ax^-\-a*x  +  x^        .         5ay-{-5xy 
Ans,  -^ T—-. -.,  -T ;—- 1 .,  and 


a*— <3u;*-fa*a;— a;''  a'— aa;*4-a*a;— a;"'         a^  -aa;*-f  a*a;— «'* 

7 


98  ALGEBRAIC  FRACTlONaf. 

PROBLEM. 

(152.)  To  reduce  an  entire  quantity  to  a  fractional  form  having 
a  given  denominator, 

RULE. 

Multiply  the  entire  quantity  hy  the  given  denominator^  and  place  it 
over  the  denominator, 

PROBLEM* 

(153.)  Reduce  1  +  a;  to  a  fraction  with  l—x  for  its  denominator, 

SOLUTION, 

Multiplying  l+ar  by  l—x,  and  placing  the  product  over  l—x,  we 

1— a;" 

have for  the  required  form. 

1— a?  ^ 

EXAMPLES, 

!•  Reduce  2  to  a  fraction,  having  4  for  its  denominator. 

Ans,  {• 

2»  Reduce  a  to  a  fraction,  having  5  for  its  denominator. 

.       ab 

Ans.  -7-r 
6 

3«  Reduce  2xy*  to  a  fraction,  having  a;'  ior  its  denominator. 

Ans.  — f-, 

X* 

4,  Reduce  a^+ab  +  6"  to  a  fraction,  having  a—b  for  its  denominator, 

Ans.  =-, 

a—o 

5»  Reduce  a*+a^x^  +  x*  to  a  fraction,  having  a^—x'  for  its  deno- 
minator. Ans. 


a'-x' 


6*  Reduce  a-f  5  +  c  to  a  fraction,  having  a  +  6— c  for  its  denomina- 

{a  +  by-c' 
tor.  Ans.  '        . 

a  +  o — c 

PROBLEM. 
(154.)  To  convert  a  fraction  into  an  equivalent  one  having  a 
given  denominator. 


ADDITION  OP  FKACTIONS.  99 

RULE. 

Divide  the  given  denominator  by  the  denominator  of  the  frojcticmy 
and  multiply  both  terms  of  the  fraction  by  the  quotient, 

PROBLEM. 

(155.)  Convert  -  into  a  fraction,  having  ab  for  its  denominator. 

SOLUTION. 

Dividing  ab  by  6,  and  multiplying  both  terms  of  ^  by  the  quotient  a, 


gives  — =  for  the  required  form. 


EXAMPLES. 

1*  Convert  |  into  a  fraction,  having  6  for  its  denominator. 

Am.  ^. 

2.  Convert  f  into  a  fraction,  having  32  for  its  denominator. 

Ans.  f|. 

3i  Convert into  a  fraction,  having  x^—y^  for  its  denominator. 

Ans.  ^'. 

4.  Convert  —5 ; — rr  into  a  fraction  having  a*  4-  6"  for  its  denom- 

a^—ab-\-h^  ® 

inator.  Ans,  -^ — ;V« 

a'  +  ft 
a^-\-x* 

5.  Convert  -7 r-r 7  into  a  fraction  having  a'— a;"  for  its  de- 

nominator.  ^tw. 


a»-aj« 


ADDITION    OF    FRACTIONS. 

PROBLEM. 

(156.)  To  find  the  sum  of  two  or  more  algebraic  fractions. 

RULE. 

Reduce  the  fractions  to  a,  common  denominator^  and  add  together 
the  numerators^  and  write  the  sum  over  the  common  denominator. 


100  ADDITION  OF  FBACTIONS. 


PROBLEM. 


(157.)  Add  — T-,  — ,  and  ■=  together. 
2o      o  7 


SOLUTION. 

Performing  the  operations  indicated  by  the  rule,  we  have 
Sa'      2a     b_105a^'  28ab      106''_105a'+28a6  +  10y 

'2b"^T'^1~~10b       ~lOb"^'lOb~  10b 

EXAMPLES. 

-     ,^^a   2a        ,  56  ,      ,,  .       20a''  +  156» 

^-  ^^^  V  Sb^  ^^^  4^  ^^'^^'''  ^^'-  -l2^6~  • 

a  +  b           a—b           ,  ^       2(a''  +  6') 

2.  Add r  and  -— y"  together.  -4«s.  -^^^ — =j^. 

3.  Add  T  and  ^  together.  Ans,  — ^^ . 
4*  Add and together.  Ans.  ~z — -. 

5.  Add  -,  -,  and  j  together.  -4w«.  a:+— . 

6.  Add  4a,  5a+— -,  and  a— ^  together.  ^w«.  10a——. 


7.  Add  3c+— -, ,  and  ^together. 

5    a—x  a      ° 

Ans.  3C+24 


Sa'a;— Sor'  +  Sa:* 


5a  (a— a;) 


x—2  2a?— 3 

8.  Add  Ix-i — -—  and  dx —  together. 

o  ^x 


A       -.«        5a:'— 16a;+9 


16a; 

A     .,,a4-ar       ,      a—x^       .  .         4ax 

9«  Add and together.  Ans.  -= -. 

a—x              a-\-x      "  a'— a;' 

-^     *  ,,           2x        ^  5x      ^             .  .             llx 

lOt  Add  5a: — =-  and  — — 4a:  together.  Ans.  a:+— — . 

7           y  oo 

11.  Add and  together.  Ans.  2-\ — =- . 

a—x              a         °  a^—ax 


SUBTRACTION  OF  FRACTIONS.  101 

11  2 

12.  Add and together.  Ans, ,, 

13.  Add ^  and  -——^  together.  Ans,  -^ f, 

1 — X  1 -7-2?  1 — X 

14.  Add  together  y,  — ,  -,  and  -.    Ans. —^ . 

15.  Add  -^„  -^,  and  -i-  together.    Ans.  2^+^y-y-+y 

a  ,    y — Qamy^ 

16.  Add  together  ^^^^^^^  and  ^^^y._^y 

^''^-    {zmf-xy  • 

-«     .,,  ,       2x    1x       ,  2a;  +  l  .       «       1.49a; 

17.  Add  together  — ,  — ,  and  — - — .  Ans,  2a?+-+— -. 

a'  6  ,  ah 

18.  Add  together  ^^^,  ^^,  and  -^^^p^. 

(a +6)' 

--..,,  ,  3a  2a+ir  ,  5 

19.  Add  together  -. — ^,  -7 — —^7 r-r,  and ; — . 

^  (a— 2a;)'''    (a  4- a?) (a --2a;)'  a+a? 

2(10g-lla;)a; 
'  (a + a;)  (a— 2a;)'* 

3  3  1^  1-a; 

20.  Add  together  ttt— rj,  ^tt— r,  ^77— r,  and 


4(1 -a;)"'  8(1 -a;)'  8(H-a;)'  4(l+a;») 

.       l+aj+a;" 
Ans, 


l_a;~a;*+a;» 


SUBTRACTION    OF    FRACTIONS. 
PROBLEM. 
(158.)  To  find  the  difference  of  two  algebraic  fractions. 

RULE. 

Reduce  the  fractions  to  a  common  denominator^  and  subtract  the 
numerator  of  the  fraction  to  he  subtracted  from  the  numerator  of  the 
other  fraction,  and  place  the  result  over  the  common  denominator. 


102  SUBTBACTION  OF   FRACTIONS. 


P&OBLBM. 


(159*)  Subtract from  -— — . 


SOLUTION. 


Performing  the  operation  indicated  in  the  rule,  we  have 

l--2a;4-a;'     l+2a;  +  g'_— 4a;  _       4a; 
1-a;"  l-a;»     T^'"     \^^' 


EXAMPLES^ 

!•  Erom  t  subtract  ^.  .4»«. 


^    _,         a-\-h      -^     ^  a— 6  .         4a6 

Z*  Prom  ^  subtract r«  ^w«.  -^ — rr, 

a— 0                  a  +  6  a  —0 

«    „         1+3;"^^^  l-a;"  .         4a;' 

3«  From  :; z  subtract  , — -5.  Am, -.. 

1— a;'                 1+a^  1—x* 

4«  From  subtract .  Ans,  -z — r. 

a—x                a-\-x  or— or 

*'    -TL.       «    .  ^      IX      X        ^— «  ^      «      ca;+&a;— aft 

Ot  From  3a; 4--  subtract  x .  Ans,  2a;  H j . 

Q                         c  be 

-    „         r     .   ^—2       ,x      X   ^       2a;— 3         .  11a;— 19 

6*  From  5a;  H —  subtract  4a; — .     Am.  x-\ , 

3                                    6  16 

7«  From  —. — -^+a  subtract  -7 r.  Am,  a- 


a(a+x)  a(a—xy  '         a'— a;** 

Q    t:^.        o        v*      ^3a  +  12a;  .       3a;— 3a 

8*  From  3a;  subtract .  Ans.  , 

5  5« 

16a; +  23 


9*  From  2a;-J 3^  subtract  8a; ^—,  Am, 


42 


lUt  rrom  ax-{ —  subtract  x — — .     Am,  ax , 

11,  From  -^  subtract  -^.  Ans,  x+—, 

7  6  35 


MULTIPLICATION  OF  FRACTIONS.  103 

1  +  2?/  6v 1 

12,  From  9y  subtract  — — ^.  Ans,  Sy+~ — . 

8  8 

13«  From  subtract ,  Ans. 


x—3  X  x^—3x 

1,1  .26 


14*  From r- subtract =•,  Ans.  tz =.  ^   ^-r     . 

Id*  From  - -\ subtract  -  H — . 

on  9.     y 

anqy-\-mbqy^pbny-'Xhnq 
bnqy 

16*  From  — ; — I — ; —  subtract .  Ans  0. 

ab         be  ae  . 

17.  From  — !-^  subtract  — -—  +  -^ ^.  .^tw.  1. 

y  x-\ry     xy—y'' 

18.  IYoml^^+^  subtract  ,^+^..  ^.      ^ 


3(l-y)     1-y                3(l-y)  1-y 

19.  From subtract  ^,    .  ^.  +777-rTT\-  -^^-  4—,  • 

a;— 1                2(a;  +  l)     2(ir''4-l)  «*— 1 

1       1       rc-1 


20.  From  — +-3+-m  subtract  -  + 7-7— t\5- 


uln^. 


x\x'  +  lf 


1^  .  ■  »  > .  »■ 


MULTIPLICATION    OF    FRACTIONS. 

PBOBLEM. 
(160.)  To  find  the  product  of  two  or  more  algebraic  fractions. 

RULE. 

Multiply  the  numerators  together  for  a  new  numerator,  and  the  de- 
nominators together  for  a  new  denominator. 


104  MULTIPLICATION  OF  FRACTIONS. 

DEMONSTRATION. 

A  C 

Let  -^  and  -yr  represent  any  two  fractions. 

•«r  ,.  ^      A        C      AC 

We  are  here  to  prove  that  -^  ^'f)—~WjT' 

A  C  A      C 

Since  —==AB-'  and  -^=  CD-\  -^  x  -^  must  equal  AB-'  x  (7i>-*. 

But  by  Prop.  1  in  Multiplication,  we  have  AR-^  x  CI)-'=AB-^ 

AB~^  CD~^ 
CD~^= which,  freed  from  negative  exponents  by  Prop, 

AC 
(101),  is  ^.    Q.E.D. 

PROBLEM. 

(16 1.)  find  the  product  of -r — and  -5 — r- — . 

SOLUTION. 

To  save  labor,  let  us  resolve  the  terms  in  both  of  the  fractions  into 
their  simplest  factors,  and  proceed  as  directed  by  the  rule,  merely  in- 
dicating the  multiplication,  and  then  canceling  the  fectors  ic+l,  jc+l, 
and  a; +  4,  common  to  both  terms,  we  have 

(ar+l)(a;  +  2)(a;+l)(a!4-4)_a:  +  2 
(«+  l)(a;+  l)(a;  +  3)(a;  +  4)~'a:+3" 

EX  AMP  LE  S. 

1«  Find  the  product  of—-  and  — -  Ans,  — , 

^  6  2a  5a 

2.  Find  the  product  of — , ,  and  — ~,  Ans,  15ax, 

^  a      c  2b 

3*  Find  the  product  of  -^,  — ,  and  — ; — .  Ans 


3 '    7  '         a-rx'  '  21(a-f  a?)* 

-    _    ,   ,  ,         /.        ar-i- 1        ^x—1  ,       Sarfx'— 1) 

i,  Fmd  the  product  of  3a;,  -— — ,  and 7.       Ans.  -—^7 -/. 

^  '    2a  '         a-\-b  2a{a  +  b) 

5*  Find  the  product  of  2aH and  3a . 

^  a  ax 

Ans.  Qd'-hSbx r. 

X         Q^ 


MULTIPLICATION  OF  FRACTIONS.  106 

6t  Find  the  product  of -,-, -„,  and  a-\- 

^  a  +  h    ax  +  x^  < 


ax 


a—x 

a'(a—h\ 
Ans.  — ^ /. 

X 

It  Find  the  product  of  -z — i-r  and r.  Ans.  -, — -7^,. 

^  a^—b^         a+b  {a  +  o) 

«    ^.    ,    ,  ,  ^a;'— 9:r  +  20       ,  a;'— 13a? +  42 

8.  Find  the  product  of 5 — and 5 — . 

X  — \)X  X  — ox 

a;'— 11a; +  28 


Ans. 


X' 


-X    ^.    ,   ,  ,         ^a—a;        ,    Va;'  ,       1(a-hx)x^ 

9*  Find  the  product  of  — - —  and .  Ans.  -^  ^    '    . 

^  3a  a^x  Za 

X  X  (2a  +  x\x 

10.  Find  the  product  of  a  +  - and  a—-  Ans.  a^—- — r— -- . 

00  15 

^-    ^.    ,   ,  ,         ^     3a;'  ,  15a;— 30        ',        ,     .x 

11.  Find  the  product  of — -  and  — Ans.  4a;+-. 

oX'~~  x\J  JdX  i 

-«.    ^.    ,   ,  ,         „2a—2x      ,     3aa;  "         .        2x 

12.  Find  the  product  of  -—- = —  and —.  Ans,  —r, 

'  3ao  5a— 5a;  66 

13.  Find  the  product  of 7-, >  and «-. 

a  +  o     a  +  x  ax — x 

.       a(a—li) 

Ans.  -i ^. 

_  X 

a* "—  a;*    a  A- it  a  — —  ti 

14.  Find  the  product  of  -5 5,  — „ =,  and .        Ans.  a  +  x, 

^  a^—7/^a^-^x*  a—x 

ax  ax  ax 

15.  Find  the  product  of  a-! and  x .         Ans.  -= z. 

*  a—x  a+x  a—x^ 

x^—h^  3;"+ 6'  a;*— 6* 

16.  Find  the  product  of  — 7--  and  -7 .  Ans.  r— ^ — r, 

OC  u  •\-  C  OC\0  +  Cj 

*w    T^.   J  XT.  1     X    i.5a     13c       Qh      .  ,       ,3a 

17.  Find  the  product  of -= --; r-+ '«  and  — ;. 

^  6       2a      bog  5d 

3a'     39ac       18aA      21a 
^*  'bd  ~~10c?'  ~25^^'^~6~* 

tQ    T?;  A,i.         A    .    f^'      ab        b'       ,  3a''     2a6  ,    b' 

18.  Find  the  product  of  —i—- h— ^  and  — 1 — ;. 

*  X*     2xy      y  a;'      5a;y      y 

3a*     19a^5     21a'6'      9a6'      ^ 
a;*      T0^"*"^5^~T0^«"*"y 


106  DIVISION  OF  FRACTIONS. 

19.  Find  the  product  of  -^^ ^ and  ^r_^y 

Ans.  2(a +  «)  +  }. 

,  ia"— 166'       ,  5a 

20.  Find  the  product  of     ^_^^     and  20a'  +  80a6  +  806»- 

Ans, 


a  +  2b' 


DIVISION    OF    FRACTIONS. 

PROBLEM. 
(162.)  To  divide  one  fraction  by  another. 

RULE. 
Invert  the  terms  of  the  divisor,  and  proceed  as  in  Multiplication, 

DEMONSTRATION. 

A  0 

Let  -^  and  -^  represent  any  two  fractions. 

^  ^     A      0    A     I)      AD 

We  are  to  prove  that  "^-^77 ="^  ^  ~r~~Wn' 

A  G  A      C 

Since  ^^AB-^  and  -^=CD-\  -g-^-^  must  equal  AB-^-r- 

CD~^,  or  ,  which,  freed  from  negative  exponents,  is  -577- 

e.  ^.  i>. 

PROBLEM. 

(163.)  Divide  ^-^  by—. 

SOLUTION. 

By  the  rule,  we  have 

2a;'       x-\-a_2x''{x-\-a)_         2x'{x+a)  _        2x 

a'+a;'        x    ~x(a^-{-x^)~x{a-\-x)(a^—ax-^x^)~a^—ax-\'a^' 

EXAMPLE  s. 

<    -rv.  .,     a;-}-a     ,      a;  +  6  .       5x^-{-Qax-{-a* 

!■  I^vide  2^-24  by  j-^^.  Jns.  -^^^_^ 

2.  Divide  ^^7  ~E''  ^^*'  ^^* 

8         9 


DIVISION  OF  FEACTIONS.  107 

3*  Divide  — - —  by  -— .  Am,  -^ \ 

4»  Divide by .  Ans.  2x, 

5  ''       6x 

Cb^'~'X^  CL      X 

5»  Divide by  — — r.  Ans.  a^  +  2a*x  +  2ax'-\-x\ 

a-\-x     ''  a*  +  2ax-\-x^ 

^    ^.  .-,    a     c  ,     e     a  .       (ad-\-bc)fh 

«    ^.  .,    x-\-a  ,      x  +  b  ,        6x^  +  Qax-\-a* 

7.  Divide r  by  — — -  Ans,   ^ — 75 . 

x—b     -^  5x  +  a  x^—¥ 

_    _.  .,    2ax-^x\      X  .           2a  +  x 

8.  Divide  —5 r  by Ans,  -3- -— ,. 

a            b                o,  b 

9.  Divide  —-7-+ r  by  —r -:,  Ans,  1. 

a  +  b      a^b     ''   a--b  a+b 

10.  Divide    ,     o/.    ■  >.a  by  ^.  Ans,  x+—. 

x^  —  2bx-\-b^     "^      x—b  x 

11.  Divide — ; —  byaH .  Ans,     \,    ,    /. 

«A    Tx-  -J    3a— 3a; ,      5a— 5x  .       3 

12.  Divide -r-  by  — -r--  Ans.  -. 

a  +  6      •'     a-\-b  5 

18.  Divide  12  by  i^+^-a.  ^w.  -j-i^. 

•^        a?  a  +ax-\-x 

^M    Tx.  .,    a*-2aV+a;*        a'— rr'  .       a'-a:" 

14.  Divide T—r — i —  by  -7-7—5.  ^           Ans.  . 

a^x+ax         ''   a*  +  x^  ax 

15.  Divide  H — ^  by  1 -.  Ans,  n. 

n-\-l    ''        n  +  \ 

«A    -i^.  .1  2ax—l  .      x—a 

16.  Divide  a-\ 7 —  by 


ax-^1 

a\bx  +  2a;')  +  a{x  +  b)  —  l 

Ans. 77 -r  . 

b{x—a) 

17.  Divide  — ^  + by —.  Ans.  — . 

a—xa  +  x    ^  a—x     a+x  2ax 

a'—h'       ^      a^  +  ah  +  b^  .         ,  ,  ,  3 

18.  Divide    ,    ^  .     „  by  7^    »  ^^.  «'  +  *'. 


108  MISCELLA]?TEOUS    PROPOSITIONS. 

«A    T^.  .,    «'c     a*c     la^c     3a*c     a\^     2aV         ,3    ,         a      3a* 
19.  Drnde^^+jr— j^-- jr+-^— J— «V   by    -^,  +  -^, 

+c*.  Arts,  Y 1 — "^• 


MISCELLANEOUS    PROPOSITIONS. 

PROPOSITION 

(164.)  1.  If  the  same  quantity  be  added  to  both  terms  of  a 
fraction  the  resulting  fraction  will  be  greater  or  less  than  the  given 
fraction,  according  as  the  numerator  of  the  given  fraction  is  less  or 
greater  than  the  denominator. 

DEMONSTBATION 
To  be  supplied  by  the  student. 

PROPOSITION 

(165.)  2.  If  the  same  quantity  be  subtracted  from  both  terms  of 
a  fraction,  the  resulting  fraction  will  be  less  or  greater  than  the  given 
fraction,  according  as  the  numerator  of  the  given  fraction  is  less  or 
greater  than  the  denominator. 

DEMONSTBATION 

To  be  supplied  by  the  student. 

PROPOSITION 

(166.)  3.  If  two  fractions  added  together  equal  unity,  their  dif- 
ference is  equal  to  the  difference  of  their  squares. 

DEMONSTRATION 

To  be  supplied  by  the  student. 

PROPOSITION 

(167#)  4.  The  sum  or  difference  of  two  quantities  divided  by 
their  product,  is  equal  to  the  sum  or  difference  of  their  reciprocals. 

DEMONSTRATION 
To  be  supplied  by  the  student. 


MISCELLANEOUS    PROPOSITIONS.  109 

PROPOSITION 

X 

(168.)  5.  If  the  difference  of  two  quantities  is  equal  to-,  their 

sum  multiplied  by  x  equals  the  difference  of  their  squares  multiplied 
byy. 

DEMONSTRATION 

To  be  supplied  by  the  student. 

PROPOSITION 
(169*)  6.  Zero  divided  by  a  finite  quantity  equals  zero. 

DEMONSTRATION. 
Let  b  represent  any  finite  quantity. 
We  are  to  prove  that  r=0» 

It  is  evident  that  the  value  of  the  expression  -  will  become  less  by 

diminishing  a  when  b  remains  constant.  Therefore,  when  a  is  as- 
sumed to  be  less  than  any  assignable  quantity,  the  value  of  the  ex- 
pression -  must  also  be  less  than  any  assignable  quantity,  or  in  other 

words,  when  a=0,  we  have  t=0.     Q.B,D. 

PROPOSITION 
(170.)  Y.  A  finite  quantity  divided  by  zero  equals  infinity. 

DEMONSTRATION. 
Let  a  represent  any  finite  quantity. 

We  are  to  prove  that  ^=QO  . 

It  is  evident  that  the  value  of  the  expression  -  will  become  greater 

by  diminishing  6,  when  a  remains  constant.  Therefore,  when  b  is 
assumed  to  be  less  than  any  assignable  quantity,  the  value  of  the  ex- 
pression T  must  be  greater  than  any  assignable  quantity,  or  in  other 

words,  when  6=0,  we  have  -=  oo,      Q.  JEJ.  D, 


110  VANISHING  FRACTIONS. 

PBOPOSITION 

(171,)  8.  Zero  divided  by  zero,  considered  without  reference  to 
the  expression  from  which  it  is  derived,  may  represent  any  value 
whatever. 

DEMONSTRATION. 

Since,  a  x  0=0,  by  making  the  last  zero  the  dividend,  and  the  other 
the  divisor,  we  have 

0 

0=" 
Since  a  may  be  of  any  value  whatever,  the  truth  of  the  proposition 
is  estabhshed. 


VANISHING    FRACTIONS. 

(172»)  A  vanishing  fraction  is  one  which  becomes  equal  to -r 
when  certain  suppositions  are  made. 

PROBLEM. 
(173.)  To  find  the  value  of  a  vanishing  fraction. 

SOLUTION. 

Let —  be  a  fraction  whose  value  is  sought  when  it  becomes 

equal  to  -.    By  (112),  we  have 

^z^x'^'  +  x^^y  +  x^Y ■\'X'y'^  -\-xy^^  ^y'^K 

x—y 

If  now  we  make  y=a;,  we  have 

^  ^  — =a;^^  +  a;"*-'  +  x"^^ +  .r™-'  +  a;*^'  +  a;*^'. 

x—x 

Since  there  must  be  m  terms  in  the  quotient,  each  of  which,  equals 

af*~\  the  whole  quotient  must  be  ma;"*-\     Hence,  we  have 

^=-=ma;"^\  when  y=^x. 

x—y      0 

In  this  example  it  may  be  seen  that  we  first  obtained  an  expression 

for  the  value  of  the  given  fraction,  before  making  the  supposition 


VANISHING  FRACTIONS.  Ill 

which  reduced  it  to  a    vanishing  fraction,  and  it  was  this  process 

that  enabled  us  to  find  the  value  of  the  given  fraction  when  it  be- 

0 

came  -. 

0 

Hence,  to  find  the  value  of  a  vanishing  fraction  the  following 


RULE. 
Mnd  an  expression  for  the,  value  of  the  given  fraction^  and  then 
make  the  supposition  necessary  to  reduce  the  given  fraction  to-. 

EXAMPLES. 

2         1.8 

!•  Find  the  value  of —  when  6= a.  Ans,  2a. 

a—o 

x^—a^ 

2*  Find  the  value  of when  xz=.a,  Ans,  3a*. 

x—a 

x*—a* 

3*  Find  the  value  of when  x=:a,  Ans,  4a'. 

x—a 

x*—a* 
it  Find  the  value  of  -; — -  when  x=a.  Ans,  2a" 

x^—a^ 

/p3  Q^  Of, 

5*  Find  the  value  of  —^ 5  when  x—a,  Ans,  — . 

x^—ar  2 

/p Q^X^  1 

6t  Find  the  value  of when  x-=^a,  Ans,  -. 

x—a  2 

x^—d^ 

7f  Find  the  value  of  , r^  when  x^^a.  Ans,  oo  . 

{x-ay 

8t  Find  the  value  of  —^ — 4"  when  xz=:a,  Ans,  0, 

a;^— a' 

9.  Find  the  value  of  -^^ ~  when  x=a.  Ans.  (2a)f , 

(x—a)^ 

X — a;' 

10.  Find  the  value  of  , when  x=l,  Ans,  4. 

1—x 

x^     cc"* 

11.  Find  the  value  of when  x=a,  Ans.  ma'^K 

x—a 

I /p» 

12.  Find  the  value  of when  x=zl.  Ans.  n. 

1—x 


I 


112  VANISHING  FRACTIONS, 

13.  Find  the  value  of  — — -— - — -— — —  when  x=z(l 

Ans,  -, 
3 

14.  Find  the  value  of — -— when  x=:^a, 

Ans.  0, 

15*  Find  the  value  of  —. — — -r — - — — -  when  xz=z±a, 

a^  —  2a'x—4ax^  +  8x^  ^ 

Ans,  00  . 

16«  Find  the  value  of :r—r-  when  x=a.  Ans,  3a, 

a — a-tx't 

17.  Find  the  value  of 7^ ^ when  x=l.      Ans,  0 

1—x^ 

18»  Find  the  value  of 7-5 — ~ — ^  when  x=a, 

{x^—a^)i 

Ans,  2^iari, 

19.  Find  the  value  of  {^-'^)^+n^'-^)^  ^^^^  ^^^ 

(x'-l)i 

Ans.  l±^i, 

20.  Find  the  value  of  )  ,     ^  .    \    \ — -f=-  when  x=zl. 

(x^—2x*  +  x^  +  x'  —  2x+l)i 

Ans.  -r- 

2* 

(Tx  -\-  cic  2cicx  n 

21.  Find  the  value  of  -^-r — -7 — r  when  x—c.  Ans.  -, 

bx^—2bcx  +  br  b 

X^  —  ttx'^  —  Cl^X  -\-  c^ 

22.  Find  the  value  of when  x=a.         Ans.  0. 


23.  Find  the  value  of  —. — -—z — r -r  when  x=a.  Ans.  go  . 

a* — 2a^x  +  2ax^ — a;*  y 

24.  Find  the  value  of ^5 —  when  a;=0.  Ans.  — . 

ic'  2a 


CHAPTER  VII. 
INVOLUTION. 

(174*)  Invohjtion  is  raising  a  given  quantity  to  a  given  power, 

PROBLEM. 
(175*)  To  raise  a  monomial  to  the  nth.  power. 

RULE. 

Multiply  all  the  exponents  contained  in  the  monomial  hy  the  index 
of  the  power,  and  prefix  +  to  the  result,  except  when  the  monomial  is 
negative  and  the  index  of  the  power  is  odd,  in  which  case  prefix  —, 

DEMONSTRATION. 

Let  Bah^c^  be  a  given  monomial  whose  nth  power  is  sought.  We 
know  by  the  definition  of  a  power  that  the  wth  power  of  Zah^c^  equals 
3a6V'  taken  n  times  as  a  factor,  or,  in  other  words,  equals  the  product 
of  3\  a\  h"^,  and  c',  each  taken  n  times  as  a  factor.      Therefore, 

{z'a'h\y=z''a\hy{cy. 

Since  h^  taken  n  times  as  a  factor  is  the  same  as  h  taken  2n  times 
as  a  factor,  we  have  (6'*)** =6'".  In  the  same  way  we  get  (c^)''=:c''*, 
whence  (3aJ'c^)'*=: S^a^i^ "^3"^  which  result  agrees  with  the  rule.  When 
a  positive  monomial  is  raised  to  any  power,  it  is  evident  that  the  result 
must  be  positive,  since,  any  number  of  positive  factors  multiplied  to- 
gether will  produce  a  positive  product ;  also,  when  a  negative  mono- 
mial is  raised  to  an  even  power  the  result  must  be  positive,  since,  an 
even  number  of  negative  factors  multiplied  together  will  produce  a 
positive  product.  But,  when  a  negative  monomial  is  raised  to  an  odd 
power,  the  result  must  be  negative,  since,  an  odd  number  of  negative 
factors  multiplied  together  will  produce  a  negative  product. 

PROBLEM 

(176.)  1.  Involve  2ai6ic'"  to  the  4th  power. 

8 


114  INVOLUTION. 

SOLUTION. 

By  the  rule,  we  have  2*a2btc*"'=iea'bc*\  Whenever  one  of  the 
factors  of  the  monomial  is  numerical,  its  power  may  be  found  arith- 
metically.    Thus,  2* =16. 

PROBLEM 

2.  Kaise  Va"*  to  the  ^th  power. 

SOLUTION. 

'PntVar=BsiTidar=A.  Since,by  (22)  and  (24),wehave  VV*= 
(a"*)'*,  we  may  write  B={A)''  or  i2=^".  By  applying  the  rule, 
we  obtain  H^  for  the  pth  power  of  jR\  and  An  for  the ^th  power  of 
-4"  ;  therefore,  i2^=:^«,  or  i2^=(^^)-«=V^^.  Since  B=Va"'= 
VA,  we  have  i?^r=(V^^  ;  but  I^=V'Ap,  therefore,  (V^^=V3^ 
which  expression,  because  A=a"'  and  AP'=q!^^^  becomes  (Va"*)^-'= 

HencBy  to  raise  a  radical  to  any  power ^  we  have  only  to  raise  the 
quantity  under  the  radical  to  that  power. 

This  principle  gives  (Vay-V'^,  {Vo^y-VaF,  (Va)'=^/a^  (Vai)' 

Since  V-4=-4",  we  have  {^/A)^=An,  If,  now,  we  suppose 2? =w, 
we  obtain  {yAY=A'^=A^=Ay  which  expression,  because  -4=a'", 
becomes  (V«'")'*=a"*. 

Hence^  to  raise  a  radical  to  h  power  equal  to  the  index  of  the  root, 
we  have  only  to  remove  the  radical. 

This  principle  gives  {yay=a,  {Vc^y^ia^  (4/^)'=— a,  (V~l)' 

=-1,  {y^y=a\  (y/I2^)''=-2%  <feo. 

PROBLEM 

3.  Raise  V— a  to  the  third  power. 

SOLUTION. 

According  to  the  first  principle  given  in  the  last  solution,  we  have 
(i^IIa)'=i^^a'  ;  but  the  cube  of  any  quantity  is  the  product  ob- 


I^rVOLUTION.  115 

tained  by  taMng  the  quantity  three  times  as  a  factor;  therefore, 
{y—af^=:V—a,^—a,^—a.  Since,  according  to  the  second  prin- 
ciple, ^—a  .1/— a,  or  (V'— «)'*=■-«,  we  have  V— a .  ^ —a .  |/— a= 
—a^—a\  whence,  we  get  (V^— a)^  =  — af^— a.  We  have  now  ob- 
tained two  expressions  for  the  cube  of  1^— a;  one  being  ^—a^^  and 
the  other  — aV— a;  hence,  in  this  particular  case,  f^— a^=—- aV— a. 
We  say  in  this  particular  case,  because  V— a^=+ay--a,  when  —a' 
has  been  obtained  by  multiplying  — a  by  a"  =+a.+(i;  for,  when 
V^'  =  Va .  a .  —a—Va\  Va  .  V— a,  we  have  V—a^z^aV—a,  since, 
according  to  the.  second  principle,  Va  ,Va=a, 

Let  us  suppose  that  a=l  in  this  problem,  or  that  we  seek  the  cube 
of  V—  1.  According  to  the  first  principle  we  have  (1^—1)^=^—1,  but 
according  to  the  second,  we  get  (1^— 1)'=  — li^— 1  =  — 4^— 1.  But 
how  can  i^—1  =  —  4^-1  ?  From  the  nature  of  the  problem,  we 
know  the  V—l  which  we  have  put  equal  to  (V—iy  is  equal  to 
V'—l'=  t^  — 1.-1.-1  =  V^.f^.V^  which  equals  —  ly^,  or 
-—4^—1,  because,  by  the  second  principlel/— l.V^— 1,  or  (f^— 1)'*,  is 
equal  to  —1. 

The  result  i^- 1  is  not  a  convenient  form  for  the  cube  of  1^—1, 
since,  the  composition  of  the  —1  is  not  apparent,  because  it  may  be 
the  product  of  —1,  +1,  and  +1,  as  well  as  —1,  —1,  and  —1. 

Hence,  we  have  4/— 1=  +i/— 1  or  —  V— 1,  according  as  the  —1 
is  considered  as  being  composed  of  the  factors  —1,  4-1,  and  +1,  or 
of  the  factors  —1,  —1,  and  —1. 

By  using  the  second  principle,  we  shall  always  arrive  at  results 
which  are  not  ambiguous.  We  may,  however,  arrive  at  equally  defi- 
nite results  by  the  first  principle  whenever  the  quantity  under  the 
radical  is  positive,  but  the  results  will  not  always  be  of  the  simplest 
form. 

EXAMPLES. 

1.  Square  3a'6.  Ans,  da^b^. 

2t  Square  lax.  Ans,  49aV. 

3.  Cube  ^a'hcK  Ans.  Qa'h'c\ 

4i  Cube  —ah^c.  Ans.  ^a^bc\ 


116      _  BINOMIAL  THEOREM. 

5.  Raise  x^y~^  to  the  4th  power.  Arts,  ar'y*. 

6.  Raise  ~a-^r~2  to  the  7th  power.  Ans, ^. 

7.  Raise  3aV  to  the  wth  power.  Ans,  3  W". 

8.  Raise  —a*"  to  the  wth  power,  n  heing  even,  Ans,  a"*'*, 
9»  Raise  — a"*  to  the  nth.  power,  n  being  odd.  Ans,  —a"*". 

10.  Raise  laT^c^  to  the  0th  power.  Ans,  1. 

1 1 .  Raise  Va  +  h  to  the  2nd  power.                ^  Ans.  a  +  h, 

12.  Raise  1/a  +  6  to  the  3rd  power.  Ans,  {a  +  h)  Va  +  b, 

13.  Raise  i^a—ic  to  the  4th  power.  Ans,  (a—xY, 

14.  Raise  aVa+x  to  the  3rd  power.  Ans,  a^+a^'x, 

15.  Raise  aVa  to  the  5th  power.  Ans,  a'Va. 

16.  Raise  —  a  1^— a  to  the  5th  power.  Ans.  —a'V^^, 

17.  Raise  f^— 1  to  the  6th  power.  Ans,  —  1. 

18.  Raise  —V—1  to  the  3rd  power,  Ans,  V—1, 
19*  Raise  —^a  to  the  3rd  power.  An^,  —aVa, 
20*  Raise  j/— a.f/— 1  to  the  3rd  power.      ajcc   -^^'  --ai^a. 


BINOMIAL   THEOREM. 

/     (177.)  The  nth  power  of  a  binomial  is  represented  by  the  follow- 
ing formula :    {x  +  y)"^=  x"" + nx'^'^y  +  n ,  —- — x'^^y^ + n .  — -— . 

ar^y*+ +n,-—-x''y'^^+xy''-^-\-y\ 

DEMONSTRATIOIT. 

A  full  demonstration  of  this  celebrated  Theorem  will  be  given  in' 
the  chapter  on  Series. 

An  inspection  of  the  above  formula  shows  that  the  exponents  of  x 
commence  with  the  index  of  the  power,  and  decrease  by  unity,  and 


BINOMIAL   THEOREM.  ,  117 

that  the  exponents  of  y  commencing  in  the  second  term  increase  by- 
unity.  Thus,  for  the  literal  part  of  the  5th  power  of  x  +  t/,  we  have 
x'  +  x*i/  +  xy  +  xY  +  xy*  +  y'. 

Let  us  now  observe  the  law  of  the  coefficients  in  the  above  formula. 

"We  see  that  the  coefficient  of  the  first  term  is  unity,  and  the  co- 
efficient of  the  second  term  is  equal  to  the  index  of  the  power,  and 
the  coefficient  of  the  third  term  is  equal  to  the  product  of  the  coeffi- 
cient of  the  second  term  by  the  quotient  obtained  by  dividing  the 
exponent  of  x  in  the  second  term  by  one  more  than  the  exponent  of 
y  in  the  second  term,  and  in  general,  any  coefficient  is  equal  to  the 
product  of  the  coefficient  of  the  preceding  term  by  the  quotient  ob- 
tained by  dividing  the  exponent  of  x  in  this  preceding  term  by  one 
more  than  the  exponent  of  y  in  the  same  term. 

Thus,  we  have  {x  +  yY=x^  +  5x^y  +  lOa^y  +  IGa^y  4- 5xy*  +  y\ 

Here  the  coefficient  of  the  third  term  is  obtained  from  5x*y  by 
dividing  the  exponent  of  a;,  which  is  4,  by  one  more  than  the  expo- 

4 

nent  of  y,  and  multiplying  the   quotient  by  5,  thus  5  .  -=10.  For 

Q 

the  coefficient  of  the  fourth  term,  we  have  10  .  -=10  ;  for  that  of  the 

3 

2  1 

fifth,  10 .  -=5 ;  and  for  that  of  the  sixth,  6  .  -=1. 
4  5 

It  may  be  observed  that  after  the  middle  term  the  coefficients 

recur  in  reverse  order ;  and  also,  that  when  the  second  term  of  the 

binomial  is  negative,  the  terms  of  any  power  of  it  will  be  alternately 

positive,  and  negative,  or  what  is  the  same,  those  terms  of  the  power 

which  contain  an  odd  power  of  y  will  be  negative. 

PROBLEM. 
(178.)  To  raise  a  binomial  to  a  given  power. 

RULE. 

Perform  the  multiplication  indicated,  or  proceed  axicording  to  the 
principles  of  the  binomial  theorem, 

PROBLEM 

(179.)  1.  Raise  (2a +  36)  to  the  6'*  power. 

SOLUTION. 

By  the  Binomial  Theorem,  we  have  (2a4-36)"'=(2CT)'  +  6(2a)'(36) 
-fl5(2a)*(36)'  +  20(2a)X35)'-(-15(2a)X36)*4-6(2a)(36)'4-(36)". 


13.8  BINOMIAL  THEOREM. 

After  performing  the  operations  indicated,  we  obtain  (2a +  36)*= 
64a'  +  5l6a'b  +  2160a'6'+4320a''6'  +  4860a'6*  +  2916a6'  4-  7296". 

PROBLEM 

2.  Cube|/2+f/3.  *     ' 

SOLUTION. 

By  the  Binomial  Theorem,  we  have  {V2+VBy=(V2y+S(V2y 

{vs)-j-3{V2){vsY+{V3y. 

After  performing  the  operations  indicated,  we  obtain  ( i/2-H/3)'= 
2^2 +  61/3+ 94/2 +  3V'3. 

But  21/2+91/2  =  11^2  and  6f 3 +  31^3=9i/3 ;  therefore,  {V2+VSy 
r^  111/2 +91/3. 

EXAMPLES. 

1 .  Raise  a  +  6  to  the  8th  power. 

Ans.  a'  +  8a'b  +  28a'6'^ +56a'6'  +  I0a*h*  +  56a'6'+28a"6''+  8a6'  + 
b\ 

2.  Raise  a— 6  to  the  9th  power. 

Ans.  a'  -  Qa'b  +  36a^6''  -  84a«6'  +  126a'6*  -  126a*6^  +  84a'6«  ~ 
SQa^b'  +  Qab'-b'. 

3.  Raise  5— 4a;  to  the  4th  power. 

Ans,  625— 2000ir  +  2400a:''— 1280a;' +  256a;*. 

4.  Raise  3— 2a;^  to  the  6th  power. 

Ans.  '729  — 2916a;'' +  4860a;*— 4320a;«  + 2160a;"— 5'76a;"  +  64a;'». 

5.  Raise  ia;  +  2y  to  the  Yth  power. 

Ans.  j^jx'  +  /^a;V  +  Y^^V  +  Y^V  +  "^^^Y  +  168a;>'  +  224a;y« 
+  128y'. 

6.  Cube  a; — .  Ans.  x' i—^3  —  )• 

X  x^       \       xl 

7.  Cube  a;'^'*-yi  Ans.  a;""*-3a;"»'*y^  +  3a;^'*y-yi 

8.  Cube  |a-|6.  Ans,  |a»-^^^6^-x««6  +  |a6«, 

^    ^  ,      a-b  ^         a'-Sa'b+Sab'-b^ 

9*  Cube -r-.  Ans,  —^ — ^  ^i  ,  in  ra — ^s* 

a— 26  a'— 6a'*6  +  12a6''— 86' 

10.  Cube  x^ + f,  Ans.  x'  +  3a;y + 3a;V'  +  y\ 

11*  Raise  a;'  +  a'  to  the  4th  power. 

Ans,  a;"  +  4a;V  +  6a;V  +  4a;V  +  a*. 


BINOMIAL  THEOEEM.  119 

12*  Cube2a-36.  Am,  8a'— 36a'6  +  64a6'— 276'. 

13*  Raise  ax-^hy  to  the  6th  power. 

Ans.  a'x'—5a'bx*j/  +  10a'b'xy—10a''b'xY  +  5ah*xy*—by. 

14.  Cube  2r— 6m.  Ans.  8r'— '72wr='  +  216wV— 216m'. 

15.  CubeV2+V5.  Ans.  Itf2  +  llf6. 

16.  CubeV^+l^"^.  Ans.  -Yj/^-Sf"^. 

17.  Raise  aVa—bVb  to  the  4th  power. 

Ans.  a'—4ayabVb+6a'b'—4aVab*Vb  +  b\ 

18.  Raise  ar-f — y~*  to  the  4th  power. 

19.  Raise  —2V^^—Sybto  the  4th  power. 

^ws.  16a''-96aV'-af^6-216a6+216i/^6V^6  +  8l5'. 

20.  Cube  (-l)i-(-l)i.  Ans.  4/-^-2. 

PROBLEM. 
(180.)  To  square  a  polynomial. 

B  U  L  E. 

Square  each  term^  and  annex  twice  its  product  irHo  each  of  the  fol- 
lowing terms. 

PROBLEM. 

(181.)  Square  2a'— 35  +  4c—5(?. 

SOLUTION. 

Squaring  2a',  we  have  4a*,  and  annexing  twice  the  product  of  2a* 
into  each  of  the  following  terms  —36,  +4c,  and  —5dj  we  obtain  4a* 
—  12a''6  +  16a'c— 20a''(?.  Also  squaring  —36,  we  have  96',  and  an- 
nexing twice  the  product  of  —36  into  each  of  the  following  terms, 
4- 4c  and  —5c?,  we  obtain  96'^- 246c4-306c?.  Also  squaring  +4c, 
we  have  16c',  and  annexing  twice  the  product  of  +4c  into  the  follow- 
in<T  term,  —Sd,  we  obtain  lec'^— 40c^.  Also  squaring  —  5(/,  we  have 
26c?^,  and  as  there  are  no  term;^  following  —5c?,  we  have  nothing  to 
annex.  Collecting  all  these  results,  we  have  (2a'— 3 6 4- 4c— 5c?)' = 
4a*-12a'^6  +  16a'c-20a'(^  +  96'-246c  +  306(?  +  16c'-40c^/  +  25c?'. 


120  BINOMIAL  THEOREM. 

EXAMPLES. 

1.  Square  a:+y+2.  Ans.  x'^-^2xy  +  2xz+y^  +  2i/z-{-z\ 

2»  Square  a—b—c  +  d. 

Ans.  a''—2ab—2ac  +  2ad-{-b^-^2bc—2hd  +  c'—2cd+d'. 

3.  Square  2a~  2  —  3aH^  +  a^b~^. 

Ans. 1263+4a6~3+9a53_6a»5+a'6~*. 

a 

4.  Square  ?y-g  +  ^-. 

,       4a*      ^      4a^d^  ,   96*       Sb'cfi  ,   d* 
^^  36^c2       **        a\^       ^ 


PROBLEM. 
(182.)  To  cube  a  polynomial. 

RULE. 

Cube  each  term^  and  annex  to  it  three  times  the  product  of  its 
square  into  ea^h  of  the  other  terms,  and  also  six  times  all  the  pro- 
ducts that  can  be  formed  by  multiplying  three  different  terms  together, 

PROBLEM. 

(183.)  Cube  a+i  +  c  +  d 

SOLUTION. 

Cubing  a  and  multipljring  three  times  its  square  into  the  other 
terms,  gives  a'  +  Za^b  +  3a^c  +  3a^d. 

Cubing  b  and  multiplying  three  times  its  square  into  the  other 
terms,  gives  6'  +  36V  +  36'c  +  36U 

Cubing  c  and  multiplying  three  times  its  square  into  the  other 
terms,  gives  c®  +  3c''a  +  3c^6  +  3c'<f . 

Cubing  d  and  multiplying  three  times  its  square  into  the  other 
terms,  gives  d^  +  3c?''a  +  3«f  6  4-  ^d'^c. 

Multiplying  by  6  all  the  products  that  can  be  formed  of  the  terms 
a,  +6,  4-c,  and  -\-d,  taking  three  at  a  time,  gives 
6a6c4-  6a6c?  +  Qacd-\-Qbcd. 


BINOMIAL  THEOREM. 


121 


Writing  all  these  results  in  order,  we  have  (a +  6 +  c +  «?)"= a' -f 

d"  +  3c?'a  +  ^d'h  +  ^dc  +  Qahc  +  Qabd  +  Qacd  +  6  W. 

Remark. — ^To  find  the  number  of  products  that  can  be  formed  of 
any  number  of  terms  taken  three  at  a  time,  add  the  numbers  of  the 
following  series  to  as  many  terras,  less  2,  as'  there  are  given  terms  : 
1+3 +  6  + 10  + 15 +  21 +  28  +  36 +  &C. 

If  we  have  seven  terms,  the  number  of  products  there  can  be  formed, 
taking  three  at  a  time,  =1+3  +  6  +  10  +  15=35. 

This  may  be  illustrated  as  follows  :     Supposing  the  seven  terms  are 
o,  6,  c,  <?,  e,  f,  and  g^  and  we  have 


ade 
adf 
adg 


Is         <l 
J  ^9  ) 


of9\l 


hde 
hdf 
bdg 


] 


leg  f 


y^fl 


cef 
ceg 


e/irh 


deg  ) 

From  these  results,  we  have 

6+  4+3+2+1 

4+  3+2+1 

3+  2  +  1 

2+  1 
1 


15  +  10  +  6  +  3  +  lJ 


or  < 


rl+2+3+  4+  6 

+  1+2+   3+   4 

1+   2+   3 

1+   2 

1 

,1+3  +  6  +  10  +  15 


"Whence,  the  law  of  the  series  becomes  apparent. 


12^  BINOMIAL  THEOKEM, 

EZAMPLBS. 

!•  Cube  a-\-b-\-c. 

Ans.  a' + Sa'b  +  3aV  +  6' + 36»a  +  SJ'c + c'  -f  3c'a  +  Sc^b  +  6aoe. 

2.  Cube  a  +  b—c—d. 

Ans.  a'  +  3a'6— Sa'^c— 3a'e?  +  6'  +  36»a-36V— 36''<?-c'  +  3c'a  + 
3c'6-3c=(;-rf'4-3c?W+3c?"6— Sf^'c— 6a5c— 6a6c?  +  6a<rrf  +  66c<?. 

3,  Cube  a  +  2b—c. 

Ans,  a'  +  ea^'fe— 3a''c4-86''  +  126'a— 126''c— c'  +  3c'a  +  6c''6-12a6c. 

4,  Cube  2x''+4ax—Sa\ 

Ans.  8a:''  +  48aa;'+60aV— 80aV— 90aV  +  106a'a?~27a'. 

5.  Cube  l+arfic'  +  a:'. 

^»«.  14-3a;  +  6a;*-|-10a;'+12ar'4-12a;'+10a;"  +  6a;'  +  3a:'+«». 

PROBLEM. 
(184.)  To  raise  a  polynomial  to  any  power. 

RULE. 

Change  the  polynomial  to  a  binomial^  and  th^n  proceed  a^icording  to 
the  principles  of  the  binomial  theorem, 

PROBLEM. 

(185.)  Find  the  6th  power  of  a  +  b-irC-\-d+e, 

SOLUTION. 

Changing  a  +  b-\-c-\-d-\-e  to  a  binomial,  we  have  (a  +  6  +  c)-f- 
{d+e)',  whence,  we  have  [(a  +  6  +  c)  +  (c?  +  e)]' =  (a  +  ^  +  c)* + 
^{a  +  b->tcy{d  +  e)  +  lb{a  +  b-\-cY{d-^ey  +  20{a  +  b^cy{d^-ey  + 
\b(a-\-b  +  c)\d  +  eY  +  Q{a  +  b  +  c){d  +  ey  +  {d  +  ey, 

In  the  same  way,  we  find 
(a4-64-c)''=:(a  +  6)"  +  6(a  +  6)'^c  +  16(a  +  6)V+20(a+6)V  + 

16(a  +  6)  V  +  Q{a  +  by  +  c\ 
(a+6-fc)^=(a+6)^  +  5(a  +  6)V  +  10(a  +  6)V  +  10(a  +  6)V4- 

6{a^by-\-c\ 
(a  +  6H-c)*  =  (a+6)*  +  4(a+&)''c  +  6(a4-6)V  +  4(o  +  6)c»  +  c*. 
(a  +  64.c)«=(a4-6)''  +  3(a  +  ft)»c  +  3(a  +  6)c»  +  c». 
\a-^b  +  cy-~  {n^by  -\-2(a  +  b)c-\'c\ 


m 


BINOMIAL  THEOEEM.  123 

Inserting  these  values  in  tlie  above  expression,  we  obtain 
(a+ft4.c4-rf  +  e)«=(a-f6)''  +  6(a  +  5)*c+15(a  +  6)V+20(a+6)V  + 
16(a  +  6)V+  6(a  +  &K  +  c«+6[(a  +  6)'+  5(a  +  6)V  +  10(a^-6)V^- 
10(a+6)V  4-  5(a  +  %*  4-  c']{d  +  e)  +  15  [(a  +  b)*  +  4(a  +  6)'c  + 
6(a+  6)V  +  4(a  +  by  +  c*]  {d+ef  +  20  [ (a  +  6y  +  3(a  +  6)'c  + 
3(a  +  bfc  +  c^]  (c^  +  e)^  4-  U[(a  +  bY  +  2(a  +  6)c  +  c»]  (c?  +  e)*  + 
6(a  +  6  +  c)(c^  +  e)^  +  (<i  +  c)«. 

EXAMPLES. 

1.  Find  the  4th  power  of  a  +  b  +  c. 

Ans,  a*  +  4a'6  4-  4a'c  4-  6a'6'  4- 1  ^a'bc  +  6aV4-4a6'  4-  I2ab*e  +  I2ahe* 
4-  4ac'  4-  6*  4-  46^c  +  66''c''  +  4bc'  +  c\ 

2.  Find  the  6th  power  of  a-i-b  +  c. 

Ans,  a"  +5a*b  +5a'c  4-10a'5'  4-20a'6c4-10aV4-10a»6'4-30a'6V 
4-30a'6c'4-10aV+5a6*4-  20afe'c4-30a6V  4-  20abc'  4-  5ac*  4-6' 
4-  56*c  4-  106V  4-  106V  +  56c*  4-  c^ 

3.  Find  the  6th  power  of  a  4-64- c. 

Ans.  a"4-6a*6  4-6aV  4-16a*6»  4-30«*6c  4-15aV4-20a'6»4-60a"6'c 
4-  60a'6c''  +  20a'c'  4-  15a'6*  4-  60a'6'c  4-  90a»6V  4-  60a»6c'  4-  16aV 
4-6a6*  4-30a6V4-60a6V  4-60a6V  4-30a6c*  4-6ac'  4-6*  4-66V 
4- 1 56V  4-  206  V  4- 1 66  V  4-  66c'  4-  c\ 

!•  Find  the  Yth  power  of  a  4-  6  4-  c. 

Ans.  a'  +  la'b-hla'c  4-21a'6''4-42a»6c4-21aV4-35a*6'4-106a*6"c 
4-105a*6c''  4-35aV  4-35a'6*  4-140a'6'c  4-210a'6V  4-140a'6c» 
4-35aV  4-21a'6'  4-105aVc  4-210a'6V  4-210a'6V  4-106a'6c* 
4-21a''c'  4-  7a6«4-42a6V4-  105a6V  4-  140a6V  4-  105a6V  4-  42a6c' 
4- 7ac' 4-6' 4- '76'c4- 216V  4- 356V  4-356V4-216V  4- '76c"  4-c'. 

5»  Find  the  4th  power  of  a4-64-c4-<^. 

6.  Find  the  6th  power  of  a4-64-c4-<^. 

Remark. — The  answer  to  the  5th  example  contains  thirtj'five  terms  and  to 
the  6th  fifty-Bix  terms. 


CHAPTER   VIII. 
EVOLUTION. 

(186.)  Evolution  is  the  reverse  of  Involution,  or  is  finding  a 
quantity  which  taken  a  certain  number  of  times  as  a  factor  will  pro- 
duce a  given  quantity. 

PEOBLEM. 
(1 87.)  To  extract  a  given  root  of  a  given  monomial. 

RULE. 

Divide  the  exponents  of  the  factors  of  which  the  monomial  is  com- 
posed by  the  index  of  the  root,  and  prefix  the  sign  of  the  monomial 
when  the  index  of  the  root  is  odd  ;  and  when  the  monomial  is  positive, 
and  the  index  of  the  root  is  even,  prefix  +  or  —  written  ±. 

DEMONSTRATION. 

Let  27a5^c'  be  a  monomial  whose  cube  root  is  sought.  If  we  re- 
solve 27a6'c"  into  three  equal  factors,  one  of  these  factors  will  be  its 
cube  root. 

Since  27=3-3-3,  and  a^=d^-a^'a^,  and  b''=b'b'h,  and  c^=c'''c'''c% 

we  have  2labY=S'S'S'a^'a^'a^'b-b'b'c^'c^-c^  z=3ahc'''3aibc^'Sahc^; 

whence,  we  see  that  Sa^bc^  is  the  cube  root  of  27a6V. 

Proceeding  according  to  the  rule,  we  shall  arrive  at  the  same  re- 
sult, for  since  2lab^c^=SWbY,  if  we  divide  each  of  the  exponents  by 

3,  we  have  S^a'^b'c^=Sahc\ 

If  the    monomial    had   been    —  27a6V,   we    would    have    had 

'^2la¥c'=  —  Sa^bc^x'—SaHc^x—3a^bc^;   whence  we  see  that  the 

cube  root  of  —2lab^c*  is  —3a^bc*,  or  that  an  odd  root  has  the  sign 
of  the  monomial. 

Also,  let  a'  be   a  monomial  whose  square  root  is  sought.     Since 


EVOLUTION.  125 

a'— +a. -f  a,  we  see  that  the  square  root  of  a*  is  +a.  But,  since 
0"=— a.— a,  we  see  that  the  square  root  of  a' may  also  be —a. 
Whence  we  derive  the  fact,  that  the  square  root  of  a  positive  quantity 
may  be  either  plus  or  minus^  which  fact  is  represented  by  the  expres- 
sion Va^  =  ±a* 

The  double  sign  ±  should  be  placed  before  the  square  root  of  a 
quantity  only  when  we  are  doubtful  in  reference  to  the  sign  it  should 
have.  For  if  we  should  be  required  to  square  +  a,  and  then  extract 
the  square  root,  we  know  that  the  only  proper  result  would  be  -fa, 
or  the  quantity  with  which  we  started ;  also,  if  we  were  required  to 
square  —a  and  then  extract  the  square  root,  we  are  equally  certain 
the  only  proper  result  would  be  —a. 

If  then,  we  have  a",  and  wish  to  extract  its  square  root,  to  give 
a  rigid  result,  we  must  know  whether  this  a^  was  obtained  by  multi- 
plying 4- a  by  +a,  or  —  a  by  —  a ;  if  the  former,  we  know  that  the 
square  root  of  this  a^  must  be  +a,  and  can  not  be  —  a ;  but  if  the 
latter,  we  know  that  the  square  root  of  this  a"  must  be  —a,  and  can 
not  be  +a.  We  see  then  that  it  is  only  proper  to  prefix  the  double 
sign  db,  when  we  are  ignorant  of  the  signs  of  the  factors  which  pro- 
duced the  quantity  under  consideration. 

PROBLEM 

(188.)  1.  Extract  the  square  root  of  —a*. 

SOLUTION. 

K  we  can  resolve  — a'  into  two  equal  factors,  both  of  which  are 
+  ,  or  both  of  which  are  — ,  then  one  of  these  equal  factors  must  be 
the  square  root  of  —a". 

But  since  neither  the  product  of  +  by  +,  nor  —  by  --  will  give 
—,  we  are  certain  that  —a"  cannot  be  resolved  into  two  equal  posi- 
tive factors,  or  into  two  equal  negative  factors.  Hence,  we  can  not 
obtain  the  square  root  of  —a^.  The  same  kind  of  reasoning  will 
show  that  we  can  not  obtain  the  even  root  of  any  negative  quantity. 

The  square  root  of  —a^  is  represented  by  V—a^^  which  expression 
may  be  simplified,  since  —a^^^a^.  —  l.  We  can  take  the  square  root 
of   a",   but  not   of  —  1  ;    whence,   we   see   that  i/— a'=V'a\— 1  = 

PROBLEM 

%  Extr^t  the  square  root  of  a. 


126  EVOLUTION. 

SOLUTION. 

Following  the  rule,  we  have  dtzai  for  the  square  root  of  a,  that  is 
♦^a=±ai.  But  we  have  gained  nothing,  since  rbai  is  merely  an- 
other mode  of  representing  that  the  square  root  of  a  is  to  be  extracted. 
Therefore,  the  formula  1/a=±ai  tells  us  nothing  more  than  that  the 
9qtLare  root  of  a  is  equal  to  the  square  root  of  a. 

If  we  were  to  assign  to  a  some  positive  numerical  value,  its  square 
root  might  be  obtained  either  exactly  or  approximately.  The  case 
would,  however,  be  different  if  we  sought  the  square  root  of  —a,  and 
should  assign  to  a  some  numerical  value,  because,  we  can  never  extract 
the  square  root  of  —1,  for  f/— a  is  equal  to  VaV—l,  the  value  of 
which  could  be  obtained  either  exactly  or  approximately  provided 
the  value  of  i^— 1  could  be  ascertained.  But  we  have  seen  that  the 
square  root  of  a  negative  quantity  can  not  be  obtained  either  exactly 
or  approximately. 

PROBLEM 

3.  Extract  the  mnth.  root  of  a. 

SOLUTION. 

To  find  the  mnth  root  of  a\  we  must  divide  the  exponent  of  a  by 
win,  which  gives  a"^,  that  is  '"Va=a»^» 

Let  us  now  take  the  »th  root  of  a*,  which  is  a^,  and  take  the  mth. 
root  of  this  result.     To  do  this,  we  must  divide  the  exponent  -  by  m, 

which  gives  a'^  for  the  with  root  of  a'^.  Hence,  we  see  that  we  arrive 
at  the  same  result  by  extracting  the  nth.  root  of  a  quantity  and  then 
extracting  the  mth  root  of  this  nth  root,  as  we  do  by  extracting  the 

mnth  root  of  the  quantity,  that  is,  '^a=\/  Va. 

The  wth  root  of  a'^  may  also  be  represented  by  \a^)  « ;  therefore, 

we  have  "'Va=\^  y/a=amn  =  [a^j  w,  or  \am)  n,      ^ 

PROBLEM 

4.  Extract  the  square  root  of  32. 

8  OLUTION. 

This  can  i)e  approximately  done  by  the  arithmetical  rule.    But  it 


EVOLUTION.  127 

is  sometimes  advisable  to  make  the  square  root  of  a  surd  depend  on 
some  smaller  surd. 

Since  32  =  16  •  2,  we  can  easily  get  the  approximate  square  root  of 
32  if  we  already  know  the  approximate  square  root  of  2,  by  multiply- 
ing the  approximate  square  root  of  2  by  ±4,  the  square  root  of  16 ; 
because  ¥32=^16-2  =Vl6V2  =  ±4V2. 


PROB  LEM 

6.  Extract  the  square  root  of  {x^  +  2x''y-\-xt/*), 

SOLUTION. 

By  factoring,  we  get  («'  +  2x^i/  +  xy"^)  =  (x^  -{-  2xy  4-  y'*)x.  We  know 
by  Theorem  1.  (109),  that  {x^-\-2xy-\-y'^)={x+yY ;  hence,  we  have 
{x^  +  2x'^y  4- xy^) •={x  +  yfx.     Extracting  the  square  root  of  (a; 4-y)'a?, 

we  get  ±:{x+y)Vx, 

EXAMPLES. 

1*  Extract  the  square  root  of  4a'' 6".  Atis.  =h2a&. 

2t  Extract  the  square  root  of  —25.  Ans.  ±5V— 1. 

3«  Extract  the  square  root  of  {a-\-hy.  Ans,  ±(a  +  6). 

4,  Extract  the  cube  root  of  64a"6'.  Ans.  4a'  Vb\ 

*  

5«  Extract  the  square  root  of  8.  Ans,  ±21/2. 

6.  Extract  the  square  root  of  24.  Ans.  ±21/6. 

7t  Extract  the  square  root  of  68.  Ans.  ±21^1 7. 

8.  Extract  the  squai-e  root  of  150.  Ans.  ±51/6. 

9.  Extract  the  fifth  root  of  — 32a*a;"y^\  Ans.  —2axY' 

10.  Extract  the  cube  root  of  —40.  Ans.  —  2V5. 

11.  Extract  the  square  root  of —16a'6 3^3.        Ans.r^^ah^d^^^c. 

833  7 

12.  Extract  the  square  root  of -.  Ans.  ± — . 

^  2057  11 

13«  Extract  the  square  root  of  4a*(a*^  +  2a6 -f- 6*). 

Ans.  ±2a'(a  +  6). 

14,  Extract  the  cube  root  of  — — r-.  An^.  — . 

128  4 


128  EVOLUTION. 

15.  Extract  the  cube  root  of  (ar'+y'  +  Sar'y  +  aay). 

Ans.  (x-\-y). 

16.  Extract  the  with  root  oi  oT'^lrPc-^d-^,  Ans,  -— . 

17.  Extract  the  fifth  root  of  —1.  Ans,  —1. 

18.  Extract  the  fourth  root  of  —a*.  Ans.  dbaf^— 1. 


19.  Find  the  value  of  %/a^  +  a^h\  Ans,  a\/\+h\ 

20.  Find  the  value  of  3^108.  Ans.  9V4. 

PROBLEM. 

(189.)  To  extract  the  wth  root  of  a  given  quantity  to  within  a 
given  fraction. 

RULE. 

Represent  the  given  quantity  in  the  form  of  a  whole  number^  and 
the  given  fraction  with  (ynefor  its  numerator^  and  then  multiply  the 
given  quantity  hy  the  nth  power  of  the  denominator,  and  extract  the 
nth  root  of  the  product  to  the  nearest  unit,  and  divide  the  result  hy  the 
denominntor  of  the  fraction. 

DEMONSTRATION. 

fl  '  c 

Let  T  he  a  quantity  whose  wth  root  is  sought  to  within  -.  Repre- 
senting -  in  the  form  of  a  whole  number,  we  have  ab~^  which  we 

/• 
shall  put  equal  to  g,  and  representing  -  in  the   form  of  a  fraction 

a 

with    1   for  its   numerator,    we    have  ^  which  becomes  —  by  put- 

'  d  m    ''  ^ 


ting  m  for 


e 

d 

c 


We  have  then  the  quantity  Q  whose  wth  root  is  sought  to  within 

— .     §=-^— — .    Let  r  be  the  root  of  the  greatest  rith  power  contained 

OmP  T^ 

in  §m" ;  then  the  value  -^-^  is  greater  than  the  value  of  — ^  and 


EVOLUTION.    .  129 

less  than  the  value  of  ^^ — ;  therefore,  the  nth  root  of  — —  or  of  Q 

r                             r"                           r  + 1 
is  greater  than  — ,  the  wth  root  of—-,  and  less  than  ,  the  »th  root 

of  -^^ '—.  or  in  other  words,  the  difference  between  the  nth  root  of 

T      ,  .  T  T -\-  1 

Q  and  — •  is  less  than  the  difference  between  —  and ;  but  the 

in  mm 

7*  7*  4"  1  1  ^ 

difference  between  —  and is  — ;  hence,  by  taking  —  for  the  nth. 

m  mm  ''  °  m 

root  of  §,  we  have  a  result  which  differs  from  the  true  result  by  less 

than  — .     Q.  E.  D. 
m 

PROBLEM. 

2 
( 1 90.)  Extract  the  square  root  of  9  to  within  -. 

SOLUTION. 

2       1 
Since  -  =77,  we  must  square  1|  and  multiply  9  by  the  result ; 

whence,  we  get  9*2^  =  201-.  We  take  either  4  or  5  for  the  square 
root  of  20J-  to  within  a  unit.  If  we  take  4  and  divide  it  by  1|,  we 
have  2 1  for  the  square  root  of  9  to  within  | ;  but  if  we  take  5  and 
divide  it  by  1^,  we  have  also  3^  for  the  square  root  of  9  to  within  |. 
In  these  two  results,  we  observe  that  the  first  is  just  \  less  than  the 
true  root,  and  the  second  ^  greater. 

Note. — In  the  following  examples  we  shall  sometimes  give  both 
results,  placing  the  most  accurate  first. 

EXAMPLES. 

1.  Extract  the  square  root  of  5  to  within  |.  Ans,  21  or  2 

2.  Extract  the  square  root  of  8  to  within  i.  Ans.  2  J  or  2|- 
3t  Extract  the  square  root  of  7  to  within  |.  Ans.  2^  or  24 
4*  Extract  the  square  root  of  f  to  within  \.  Ans.  ^ 
5»  Extract  the  square  root  of  4  to  within  -i.  Ans.  4 
6*  Extract  the  square  root  of  ^  to  within  J-.  Ans.  f 
7.  Extract  the  square  root  of  {}  to  within  ^.  Ans.  }| 


Am 

.  1. 

Arts. 

If 

Ans, 

2\. 

Ans. 

H- 

Am 

.  f 

180  •     EVOLUTION. 

8#  Extract  the  square  root  of  ^  to  within  J. 
9t  Extract  the  cube  root  of  5  to  within  J. 
10*  Extract  the  cube  root  of  10  to  within  ^. 

11.  Extract  the  fourth  root  of  7  to  within  \, 

12.  Extract  the  square  root  of  |  to  within  \. 

PROBLEM. 
(191.)  To  extract  the  square  root  of  a  polynomial. 

BULB. 

1.  Arrange  the  given  polynomial  according  to  the  powers  of  a  cer- 
tain letter. 

2.  Extract  the  square  root  of  the  first  term  of  the  polynomial,  and 
place  the  result  as  the  first  term  of  the  required  root, 

3.  Subtract  the  square  of  the  first  term  of  the  root  from  the  given 
polynomial. 

4.  Divide  the  first  term  of  the  remainder  by  twice  the  first  term  of 
the  root^  and  pla^e  the  quotient  as  the  second  term  of  the  root. 

6.  Annex  the  second  term  of  the  root  to  twice  the  first  term  of  the 
root,  and  multiply  the  sum  by  the  second  term  of  the  root,  and  sub- 
tract the  product  from  the  first  remainder. 

6.  Divide  the  first  term  of  the  second  remainder  by  twice  the  first 
term  of  the  root,  and  place  the  quotient  as  the  third  term  of  the  root. 

7.  Ann£x  the  third  term  of  the  root  to  twice  the  sum  of  the  two 
preceding  terms,  and  multiply  the  trinomial  thus  formed  by  the  third 
term  of  the  root,  and  subtract  the  product  from  the  second  remainder. 
Th&n  proceed  in  the  same  manner  to  find  the  other  terms  of  the  root, 

DEMONSTRATION. 
The  accuracy  of  this  rule  depends  upon  the  following  principles. 

PRINCIPLE 

1.  When  a  polynomial  and  its  square  are  arranged  according  to  the 
powers  of  a  certain  letter,  the  first  term  of  the  square  is  the  square  of 
the  first  term  of  the  polynomial. 

This  principle  is  only  a  particular  case  of  that  upon  which  the 
division  of  polynomials  is  based. 


EVOLUTION.  131 

PEINCIPLE 

2.  When  a  polynomial  and  its  square  are  arranged  according  to  the 
powers  of  a  certain  letter,  and  the  square  of  the  sum  of  n  terms  of  the 
polynomial  is  subtracted  from  its  complete  square,  the  fir^t  term  of  the 
remainder  is  twice  the  product  of  the  first  term  of  the  polynomial  by 
its  (n+l)st  term* 

Let  A  represent  the  first  n  terms  of  a  polynomial  and  B  the  re- 
maining terms.  Then  the  whole  polynomial  may  be  represented  by 
A  +  B,  which,  we  will  suppose,  is  arranged  according  to  the  decreas- 
ing power  of  a  certain  letter.  If  we  subtract  from  the  square  of 
A+B,  which  is  A'^  +  2AB  +  B^,  the  square  of  the  first  n  terms  of 
A  +  B,  which  is  A^,  there  reroains  2AB-\-B'^. 

It  is  evident  that  in  the  remainder  2AB-\-B^,  the  first  term  of  2AB 
contains  a  higher  power  of  the  leading  letter  than  any  of  the  other 
terms,  and  is,  therefore,  the  fii'st  term  of  the  remainder.  But  the  first 
term  of  2AB  is  twice  the  product  of  the  first  term  of  A,  which  is  the 
first  term  of  the  proposed  polynomial  by  the  first  term  of  B,  which  is 
the  (n-\-l)st  of  the  same  polynomial.  Hence,  the  principle  is  estab- 
lished. 

Let  us  now  proceed  to  the  extraction  of  the  square  root  of  a  poly- 
nomial, which,  as  well  as  its  root,  we  shall  conceive  to  be  arranged 
according  to  the  powers  of  a  certain  letter.  By  the  first  Principle, 
we  know  that  the  first  term  of  the  polynomial  is  the  square  of  the  first 
term  of  the  root ;  hence,  we  shall  obtain  the  first  term  of  the  root  by 
extracting  the  square  root  of  the  first  term  of  the  polynomial.  By 
the  second  Principle,  if  we  subtract  from  the  polynomial  the  square  of 
the  first  term  of  the  root,  the  first  term  of  the  remainder  will  be  twice 
the  product  of  the  first  term  of  the  root  by  the  second  term  of  the 
root ;  hence,  we  shall  obtain  the  second  term  of  the  root  by  dividing 
the  first  term  of  the  remainder  by  tTvice  the  first  term  of  the  root. 
Also,  by  the  same  Principle,  if  we  subtract  from  the  proposed  poly- 
nomial the  square  of  the  sum  of  the  first  two  terms  of  the  root,  the 
first  term  of  the  remainder  will  be  twice  the  product  of  the  first  term 
of  the  root  by  the  third  term  of  the  root ;  hence,  we  shall  obtain  the 
third  term  of  the  root  by  dividing  the  first  term  of  this  remainder  by 
twice  the  first  term  of  the  root. 

We  may  observe  that,  instead  of  subtracting  from  the  given  poly- 
nomial the  square  of  the  sum  of  the  first  two  terms  of  the  root,  it  will 
produce  the  same  result  to  subtract  from  the  first  remainder  twice  the 
product  of  the  first  term  of  the  root  by  the  second  t«rm  of  the  root. 


132  EVOLUTION. 

plus  the  square  of  the  second  term  of  the  root,  since,  we  have  already 
subtracted  the  square  of  the  first  term  of  the  root.  Therefore,  if  we 
write  the  second  term  of  the  root  after  twice  the  first  term  of  the 
root,  and  multiply  the  binomial  thus  formed  by  the  second  term  of 
the  root,  and  subtract  the  product  from  the  first  remainder,  and  divide 
the  first  term  of  this  new  remainder  by  twice  the  first  term  of  the 
root,  we  shall  obtain  the  third  term  of  the  root.  Continuing  thus,  we 
shall  obtain  successively  all  the  terms  of  which  the  root  is  composed, 
and  this  process  is  exactly  that  given  in  the^ule  ;  hence,  the  accuracy 
of  the  rule  is  rigidly  established. 

PROBLEM. 

(192.)  Extract  the  square  root  of  a*+a:*  +  6aV— 4a'a;— 4aa;'. 


a^ 

2a'  —  2ax)— 4a^x  +  6a^x'' —4ax^  +  x* 

2a'—4ax  +  x^  )  2aV— 4aa;'  +  a;* 
2aV— 4aa:'  +  a;* 

SOLUTION. 

1.  We  arranged  the  given  polynomial  according  to  the  decreasing 
powers  of  a. 

2.  We  extracted  the  square  of  a*,  and  placed  the  result  a^  as  the 
first  term  of  the  root. 

3.  We  subtracted  the  square  of  a'',  which  is  a*,  from  the  given 
polynomial,  and  obtained  the  remainder,  —4a^x-\-6o.'^x^—4ax^-{-x*. 

4.  We  divided  —Aa^x  by  20^,  and  placed  the  result,  —2aXj  as  the 
second  term  of  the  root. 

5.  We  annexed  —  2a^  to  2a\  thus  making  2a''— 2aa;,  and  multi- 
plied this  binomial  by  —2ax,  and  subtracted  the  product,  —ia^x-^ 
4:0^ x"^^  from  —4a^x  +  Qa'^x'^—4ax^  +  a;^  and  thus  obtained  the  remainder, 
2aV— 4aa:"+a:\ 

6.  We  divided  2a^x'^  by  2a'',  and  placed  the  result  x^  as  the  third 
term  of  the  root. 

7.  We  annexed  x^  to  twice  a'— 2aa?,  thus  making  2a''— 4air  +  a;', 
and  multiplied  this  trinomial  by  a;",  and  subtracted  the  product,  2aV 
—4ax^-\-x*^  froip  2a V  —  4arr' +  a?*,  and  obtained  no  remainder. 


EVOLUTION.  188 

Scholium. — If  we  had  arranged  the  given  polynomial  according  to 
the  decreasing  powers  of  .r,  we  should  have  obtained  for  the  root 
Q^  —  1ax-\-a^^  which  is  the  same  as  a^—2ax-\-x^^  differeotly  arranged. 
If  we  had  taken,  in  the  above  process,  —a^  as  the  root  of  a*,  the  com- 
plete result  would  have  been  —  a^  +  2aa;— a;",  or  ^ax—(a^-\-x^'). 
The  two  roots  of  the  polynomial  are  given  by  the  expression 
±.{a^—'lax^x^), 

E  XAMP  LE  S. 

!•  Extract  the  square  root  of  a' +  2a6  4- 6^  Am,  ±(a4-6). 

2.  Extract  the  square  root  of  a2«_2a'«52  +6". 

Ans,  ar—h"^  or  fta— a*". 

3i  Extract  the  square  root  of  25a;— 70ari  +  49ari-. 

Ans.  ba^k—^x^  or  Yrci— Sari-. 

4a'     \1oic      9c' 

4.  Extract  the  square  root  of  -r^ hT"^'^' 

2a     3c         3c     2a 
^^.  ___or--_. 

iC'  9  3 

5.  Extract  the  square  root  of  j^—ix^  +  ^x2 

-2        -S  as 

Arts.  ia;2_ia;4   q^  ^x*—lx^. 

6.  Extract  the  square  root  of  9a"a;^— 42a~r"ic  4    -^i9a"'X2. 

n     m  m     n  m     n  n     m 

Ans,  SaJxl—^alxl  or  ^dax'i—SazXi. 

7.  ExtractthesquarerootoflOa;*— lOa;"— 12a;''  +  9ar"'— 2a;  +  14-6a;'. 

Ans.  ±(3a;'-2a;'  +  a;— 1). 

,2a6V  +  6V  +  aV 
8»  Extract  the  square  root  of 


ax  +  6  V 


Ans. 


a'^+x^ 

9.  Extract  the  square  root  of  ix""— 20x*  a^  +  2 5x^a^  +  2 4x^1/^ b^ 
-eOx^y^ah^  +  S6by^.  Ans.  ±{2x^—5x^a^  +  6y«6^). 

lOt  Extract  the  square  root  of  a*— 2a'4-Ja'— ^a  +  Jj. 

Ans.  ±(a'-a-f-i). 


184  EVOLUTION. 

PROBLEM. 
(193.)  To  extract  the  cube  root  of  a  polynomial. 

RULE. 

1.  Arrange  the  given  'polynomial  according  to  the  powers  of  a  cer- 
tain letter, 

2.  Extract  the  cube  root  of  the  first  term  of  the  polynomial^  and 
place  the  result  as  the  first  term  of  the  required  root. 

3.  Subtract  the  cube  of  the  first  term  of  the  root  from  the  given  poly- 
nomial. 

4.  Divide  the  first  term  of  the  remainder  by  three  times  the  square 
of  the  first  term  of  the  rooty  and  pla^e  the  quotient  as  the  second  term 
of  the  root, 

6.  Cube  the  sum  of  the  first  two  terms  of  the  root,  and  subtract  the 
result  from  the  given  polynomial. 

6.  Divide  the  first  term  of  the  second  remainder  by  three  times  the 
sqicare  of  the  first  term  of  the  root,  and  place  the  results  as  the  third 
term  of  the  root. 

7.  Cube  the  sum  of  the  first  three  terms  of  the  root,  and  subtract 
the  result  from  the  given  polynomial.  Then  proceed  in  the  same  man^ 
ner  to  find  the  other  terms  of  the  root. 

DEMONSTRATION. 
The  accuracy  of  this  rule  depends  on  the  following  principles. 

PRINCIPLE 

1.  When  a  polynomial  and  its  cube  are  arranged  according  to  the 
powers  of  a  certain  letter,  the  first  term  of  the  cube  is  the  cube  of  the 
first  term  of  the  polynomial. 

This  principle  is  only  a  particular  case  of  that  upon  which  the 
division  of  polynomials  is  based. 

PRINCIPLE 

2.  When  a  polynomial  and  its  cube  are  arranged  according  to  the 
powers  of  a  certain  letter,  and  the  cube  of  the  sum  of  n  terms  of  the 
polynomial  is  subtracted  from  its  complete  cube,  the  first  term  of  the 
remainder  is  three  times  the  product  of  the  square  of  the  first  term  of 
the  polynomial  by  its  {n  +  l)st  term. 


EVOLUTION.  135 

Let  A  represent  the  first  n  terms  of  a  polynomial,  and  B  the  re- 
maining terms.  Then  the  whole  polynomial  may  be  represented  by 
A  +  B,  which  we  will  suppose  is  arranged  according  to  the  decreasing 
powers  of  a  certain  letter.  If  we  subtract  from  the  cube  of  A-^-B^ 
which  is  A''-hSA^B-{-SAB''  +  B%  the  cube  of  the  first  n  terms  of 
A^-B,  which  is  A\  there  remains  3^'^ +  3^^' 4-^'.  It  is  evident 
that  in  the  remainder  SA^B  +  SAB^-^B^  the  first  term  of  SA'^B  con- 
tains a  higher  power  of  the  leading  letter  than  any  of  the  other  terms, 
and  is  therefore  the  first  term  of  the  remainder.  But  the  first  term 
of  SA^B  is  three  times  the  product  of  the  square  of  the  first  term  of 
A,  which  is  the  first  term  of  the  proposed  polynomial,  by  the  first  term 
of  jB,  which  is  the  {n+l)st  term  of  the  same  polynomial.  Hence  the 
principle  is  established. 

Let  us  proceed  now  to  the  extraction  of  the  cube  root  of  a  poly- 
nomial which,  as  well  as  its  root,  we  shall  conceive  to  be  arranged 
according  to  the  powers  of  a  certain  letter.  By  the  first  Principle, 
we  know  that  the  first  term  of  the  polynomial  is  the  cube  of  the  first 
term  of  the  root ;.  hence,  we  shall  obtain  the  first  term  of  the  root  by 
extracting  the  cube  root  of  the  first  term  of  the  polynomial.  By  the 
second  Principle,  if  we  subtract  from  the  polynomial  the  cube  of  the 
first  term  of  the  root,  the  first  term  of  the  remainder  will  be  three 
times  the  product  of  the  square  of  the  first  term  of  the  root  by  the 
second  term  of  the  root;  hence,  we  shall  obtain  the  second  term  of 
the  root  by  dividing  the  first  term  of  the  remainder  by  three  times 
the  square  of  the  first  term  of  the  root.  Also,  by  the  same  principle, 
if  we  subtract  from  the  proposed  polynomial  the  cube  of  the  sum  of 
the  first  two  terms  of  the  root,  the  first  term  of  the  remainder  will  be 
three  times  the  product  of  the  square  of  the  first  term  of  the  root  by 
the  third  term  of  the  root ;  hence,  we  shall  obtain  the  third  term  of 
the  root  by  dividing  the  first  tenn  of  this  remainder  by  three  times 
the  square  of  the  first  term  of  the  root.  Continuing  thus,  we  shall 
obtain  all  the  terms  of  the  root.  This  process  is  exactly  that  given  in 
the  rule ;  hence,  the  accuracy  of  the  rule  is  rigidly  established. 

PROBLEM. 

(194.)   Extract  the  cube  root   of  a;'— 6aj*  +  21a;*— 44a;*  +  63a:" 

~64ar  +  27. 


136  EVOLUTION. 

Operation. 


3a;*)  —6a;' . 

;.  ,     . > 

I 


a;°— 63.-^  +  12a;"—  8  a;' 
3a;')  9a;* 


a;'— 6a;'4-21a;*— 44a;'  +  63a;'— 54a;  +  27. 

Remark. — ^The  student  should  observe  that  it  is  not  necessary  to  bring  down 
any  terms  of  the  remainder  except  the  first,  as  was  done  in  the  above  opera- 
tion. 

SOLUTION. 

1.  We  aiTanged  the  given  polynomial  according  to  the  decreasing 
powers  of  x. 

2.  We  extracted  the  cube  root  of  a;'  and  placed  the  result,  «',  as 
the  first  term  of  the  root. 

3.  We  subtracted  the  cube  of  x^  which  is  x*  from  the  given  poly- 
nomial. 

4.  We  divided  —Qx^^  the  first  term  of  the  remainder,  by  3  a;*,  or 
three  time  the  square  of  a;'*,  and  placed  the  result,  —2a;,  as  the  second 
term  of  the  root. 

5.  We  cubed  a;''— 2a;,  which  is  the  sum  of  the  first  two  terms  of  the 
root,  and  subtracted  the  result,  a;"— 6a;^  +  12a;*— 8a;',  from  the  given 
polynomial. 

6.  We  divided  9a;*,  the  first  term  of  the  second  remainder,  by  3a;*, 
or  three  times  the  square  of  a;",  and  placed  the  result,  3,  as  the  third 
term  of  the  root. 

7.  We  cubed  a;"  —  2a;  -f  3  which  is  the  sum  of  the  first  three  terms 
of  the  root  and  subtracted  the  result,  a;^— 6a;^  +  21a;*  — 44a;'  +  63a;' 
—64a; +  2  7,  from  the  given  polynomial,  which  left  no  remainder. 

EXAMPLES. 

1 .  Extract  the  cube  root  of  a'  +  3a^6  +  ^ah^  +  6'.         Ans.  a  +  h, 

2.  Extract  the  cube  root  of  8a'a;'— 84a'*fta;*  +  294a6V— 3436'a;". 

Ans.  2ax—1bx', 

3.  Extract  the  cube  root  of  8a;' —  36aa;'4-102aV  —  lYlaV 
4-204aV— 144a'a;+64a«.  Ans.  2x''-3ax^4a\ 

4.  Extract  the  cube  root  of  a;"— 9a;' +  39a;*— 99a;' +  156a;''— 144a; 
+  64.  Ans.  a;^  — 3a;  4-4. 


EVOLUTION.  137 

5.  Extract  the  cube  root  of  aj'  +  6a;*--40ir'H-96ar— 64. 

Ans.  a;'+2a;— 4. 

6.  Extract  the  cube  root  of  a;"— 6a;' +  15a;*— 20a;' + 15a;''— 6a; +  1. 

Ans,  a;''— 2a;  +  l. 

7.  Extract  the  cube  root  of  a^+Sa'b  +  3a'c  +  3a6' +  3acH  6a6c 
+b'  +  Sb''c  +  Sbc''  +  c\  Ans.  a+b  +  c, 

8.  Extract  the  cube  root  of  27a;'— 54a;''  +  63a;*— 44a;'  +  21a;'— 6a? 
+  1.  Ans.  3a;''— 2a;+l. 

9.  Extract  the  cube  root  of  {a  +  6)'  +  3(a  +  bye  +  S{a+b)c^  +  c'. 

Ans.  a  +  b+c. 
10*  Extract  the  cube  root  of  1  —  6a;  + 1 2a;* — 8a;'.        Ans.  1 — 2a?. 

PR  OBLEM. 
(195*)  To  find  the  mth  root  of  a  polynomial. 

BULE. 

1.  Arrange  the  polynomial  according  to  the  powers  of  a  certain 
letter. 

2.  Extract  the  mth  root  of  the  first  term  of  the  polynomial,  and 
place  the  result  as  the  first  term  of  the  root. 

3.  Raise  the  first  term  of  the  root  to  the  wth  power,  and  subtract  the 
result  from  the  given  polynomial. 

4.  Divide  the  first  term  of  the  remainder  by  m  times  the  (m— l)st 
power  of  the  first  term  of  the  root,  and  place  the  quotient  as  the 
second  term  of  the  root. 

5.  liaise  the  sum  of  the  first  two  terms  of  the  root  to  the  mih  power, 
and  subtract  the  result  from  the  given  polynomial. 

6.  Divide  the  first  term  of  the  second  remainder  by  m  times  the 
(m—l)st  power  of  the  first  term  of  the  root,  and  place  the  quotient 
as  the  third  term  of  the  root.  Continue  thus  until  all  the  terms  of 
the  root  are  obtained. 

DEMONSTRATION 

To  be  supplied  by  the  student. 

PROBLEM. 

(196.)  Extract  the  4th  root  of  a*— 4a''a;  +  a;*-|-6aV— 4aa;". 


188  EVOLUTION. 

Operation. 


4a'')  —4ta'x 


a4_4a^c  +  6aV— 4aa;'+a:* 


SOLUTION. 

1.  We  arranged  the  polynomial  according  to  the  decreasing  powers 
of  a. 

2.  We  extracted  the  4th  root  of  a*  and  placed  the  result,  a,  as  the 
first  term  of  the  root. 

3.  We  raised  a—x  to  the  4th  power,  and  obtained  a*— 4a'a; 
-h6aV— 4aa;'H-a;*,  which,  subtracted  from  the  given  polynomial,  left 
no  remainder. 

EXAMPLES. 

1  •  Find  the  second  root  of  x^  +  2xy + y' + Qxz  4-  Qyz  +  9z'. 

Ans.  db(ar  +  y  +  30). 

2.  Find  the  third  root  of  a»— 6aV-hl2a«''— 8a:'. 

Ans.  a—2x. 

3.  Find  the  fourth  root  of  16a*  — 96a'a;  +  216aV— 216aa;'  +  81a;\ 

Ans.  ±(2a— 3a:). 

!•  Find  the  sixth  root  of  x* — 6a:'  + 1  bx"  —  20a:'  + 1 5a:'  —  6a;+ 1 . 

Ans.  rfc(a:— 1). 

5.  Find  the  eighth  root  of  a:'  +  8a:' +28a:"  +  56a:'  +  '70a;*4-56a;' 
-f-28a:''  +  8a:  +  l.  Aris.  ±(a;+l). 


CHAPTER  IX. 
RADICALS. 

(197.)  A  radical  expression  is  one  which  contains  one  or  more 
radical  signs,  or  fractional  exponents ;  as,  Va^  ai,  Va",  <fco. 

(198.)  A  rational  quantity  is  one  which  may  be  represented 
without  the  aid  of  a  radical  expression.  Thus,  Va^^  8%  and  VSl  are 
rational  quantities,  because  they  may  be  represented  ±a,  2,  and  ±3. 

(190.)  An  irrational  or  surd  quantity  is  one  which  can  not  be 
represented  without  the  aid  of  a  radical  expression ;  as,  Va,  VS,  2^,  <fec. 

(200«)  An  imaginary  quantity  is  one  which  is  represented  by  a 
radical  expression  denoting  the  even  root  of  a  negative  quantity ;  as, 
f  ^,  V-"2,  (-4)i,  &c. 

(201,)  K  quadratic  surd  is  one  in  which  the  root  indicated  is 
the  square  root ;  as,  Va^  Vh^  1^2,  <fec. 

(202t)  A  cubic  surd  is  one  in  which  the  root  indicated  is  the 
cube  root ;  as,  Vff,  fti,  1/2,  <fec. 

THEOREM. 

(203.)  TTie  sqvxtre  root  of  a  quantity  can  not  he  partly  rational 
and  partly  a  quadratic  surd. 

DEMONSTRATION. 
K  possible,  let  the  square  root  of  a  be  b-\-Vc.  It  follows  then  that 
a  must  be  equal  to  the  square  of  b+Vc,  or  b^  +  2bVc-{-Cj  that  is,  a 
rational  quantity  is  equal  to  the  sura  of  two  rational  quantities  and  an 
irrational  one,  which  is  manifestly  impossible,  therefore,  Vaz=b-\-Ve 
represents  an  impossibility. 

THEOREM. 
(204.)  In  any  equation  consisting  of  rational  quantities  and 


140  REDUCTION  OF  SURDS. 

qiLadratic  surds,  the  sum  of  the  rational  quantities  on  each  side  of  the 
sign  of  equality  is  equal,  and  also,  the  sum  of  the  quadratic  surds, 

DEMONSTRATION. 

li a+b-\-Vc-\-Vd-\-Ve=m+n+p-\-Vr+\^p,  then  a  +  b=m  +  n-^Pf 
and  Vc+Vd+Ve=Vr+Vp.  Let  A=a  +  b  and  VB=Vc-]-Vd+V7; 
also,  C=m+n+p  andV'B=Vr-WPi  then  A+VB'=0+VJD>: 

If  A  does  not  equal  0,  let  it  equal  Cdzt,  then  C±t+VB'=  C-^VD, 
or  i^t+VBzzzVD,  which  shows  that  the  square  root  of  D  is  equal  to 
a  quantity  partly  rational  and  partly  a  quadratic  surd,  which  by  the 
last  Theorem  is  impossible,  imless  ±Ms  equal  to  nothing;  whence 
VB=VD  and  A=a     Q.  E.  D. 

THEOREM. 
(205.)  If  the  square  root  of  a+Vb  equals  x-\-y,  then  the  square 
root  ofa—Vb  equals  x—y\x  and  y  being  supposed  to  be  one  or  both 
quadratic  surds, 

DEMONSTRATION. 

By  hypothesis  a+Vbzzzx^ -{-^xy -{•y'^=x'^ -\-y'^  +  2xy.  It  is  evident 
that  x^+y"^  must  be  rational,  since  both  x^  and  y"  are  rational, 
whether  x  ot  y  are  supposed  to  be  rational  or  quadratic  surds. 
Hence,  by  the  last  Theorem,  we  must  have  a=x^+y''  andi^6==2ay, 
whence  a—Vb=a^-{-y''—2xy=x''-2xy+y'^,  or  i/a— 1^6  =a?— y. 


REDUCTION    OF    SURDS. 

PROBLEM. 

(206,)  To  reduce  a  rational  quantity  to  the  form  of  a  proposed 
surd. 

RULE. 

Involve  the  given  rational  quantity  to  a  power  denoted  by  the  index 
of  the  proposed  surd,  and  then  represent  the  corresponding  root  of  the 
result  by  means  of  a  radical  sign  or  fractional  exponent. 


EEBUCTION  OF  SUBDS.  141 

PROBLEM. 

(207.)  Reduce  a  to  the  form  of  a  cubic  surd. 

SOLUTION. 

The  cube  root  of  a'  or  VaF={a^)^  is  evidently  the  required  form. 

EXAMPLES. 

!•  Reduce  3  to  the  form  of  a  quadratic  surd.        Ans.  V9  or  9^. 
2.  Reduce  3a;'  to  the  form  of  a  cubic  surd.  Ans,  V^lx', 

3*  Reduce ; to  the  form  of  a  cubic  surd. 

a+b+x  

^  (a  +  b+it)'' 
I.  Reduce  a+a;  to  the  form  of  a  quadratic  surd. 


Ans 

\ya'  +  2ax+x\ 

5. 

V2 
Reduce  — -  to  the  form  of  a  quadratic  surd. 
6 

Ans.  /A. 

6. 

Reduce  -^  to  the  form  of  a  cubic  sui'd. 

V4r 

AnsYf. 

7. 

Reduce  a*  6*  to  the  form  of  the  fifth  root. 

Ans.  (ab')i. 

8.  Reduce  Va  to  the  form  of  a  quadratic  surd.       Ans,  JuVa^ 
9i  Reduce  —a  to  the  form  of  a  quadratic  surd. 


Ans,  Va^=.V{-d)\ 
10»  Reduce  —a  to  the  form  of  the  fourth  root.       Ans,  V(— a)*. 

PROBLEM. 

(208»)  Reduce  a^h  to  the  form  of  a  quadratic  surd. 

SOLUTION. 

Since  a=f/a^,  we  have  (]f^h-=S/ayh=S/€Fh, 

EXAMPLES. 

1,  Reduce  2V3  to  the  form  of  a  quadratic  surd.  Ans.  VVl, 

2«  Reduce  3\/2  to  the  form  of  a  cubic  surd.  Ans.  1/54, 

3*  Reduce  a\fh  to  the  form  of  the  mth  root.  Ans,  "^JcS^h, 


142  REDUCTION  OF  StJEDS. 


4,  Reduce  (a—b)Va^  +  b'*  +  2ab  to  the  form  of  a  quadratic  surd. 


Ans.  Va'  —  2a'b^-^b\ 

PBOBLEM. 

(209.)  To  reduce  two  or  more  radicals  having  different  indices 
to  equivalent  ones  having  the  same  index. 

RULE. 

Represent  the  given  radicals  by  the  aid  of  fractional  exponentSj  and 
reduce  these  fractional  exponents  to  equivalent  ones  having  a  common 
denominator  ;  then  raise  ea^h  quantity  respectively  to  the  powers  de- 
noted by  the  numerators  of  these  fractions,  and  the  common  denomina- 
tor will  be  the  index  of  the  root  of  each, 

PRO  BLEM. 

(210.)  To  reduce  2\/3  and  34^2  to  surds  expressing  the  same 
root 

SOLUTION. 

We  have  2l/3=V24  =  (24)3  and  3f  2=Vl8=:(18)i.     But  i=f 

and|=^;  whence,  (24)i=(24)«=(24'')»=576^=V676,and  (18)^ 

=(18)°=(18')e=(5832)«=:V6832.      Therefore,  2V3=V576  and 
3f^2=V5832. 

EXAMPLES. 

1«  Reduce  V2^and  V4  to  surds  expressing  the  same  root. 

Ans.  V8  and  VI 6. 

2*  Reduce  Vx^  and  Vy  to  surds  expressing  the  same  root. 

Ans,  V"^'  and  "v//. 

3.  Reduce  Vax  and  X/ba^  to  surds  having  a  common  index. 

Ans,  VaV  and  V6V. 

4i  Reduce  ^3  and  \/2  to  surds  having  a  common  index. 

Ans,  V27  and  V4. 

5^  Reduce  6^  and  5*  to  surds  having  a  common  index. 

Ans.  Ve^and  V5^ 


REDUCTION  OF  SURDS.  143 

6*  Reduce  2*,  3^,  and  f  5  to  surds  having  a  common  index. 

Ans.  ^512,  V6561,  and  V15625. 


7»  Reduce  aVa—x  and  bVo,^—x^  to  surds  having  a  common  index. 
Ans.  l/a'-3a'a;H-3aV— aV  and  Va*b'  —  2a'b'x^  +  b'x\ 
b 


8,  Reduce  aVx—y  and  T7==  to  surds  having  a  common  index. 
^»«.  VaV— 2a*a:y+ay  and  V^X^Typ'. 


9t  Reduce  Va^—x^  and  Va*+ic*  to  the  form  of  the  eighth  root. 
Ans.  X/ia^'—xy  and  l/a'  +  2aV+?. 

lOi  Reduce  {a  +  x)\  and  (a— a;)f  to  surds  having  a  common  index. 
Ans,  (o'  +  3a'a;  +  3aa;'  +  a;')i  and  (a*— 4a'a;  +  6aV— 4aa;'+a?*)^. 

PROBLEM. 

(21  !•)  To  reduce  surds  to  their  simplest  form. 

BULB. 

Separate  the  quantity  under  the  radical  into  two  factors^  one  of 
which  must  be  the  greatest  perfect  power  corresponding  to  the  root 
indicated  that  is  contained  in  the  given  quantity.  Extract  the  root  of 
this  factor^  and  place  the  product  of  it  by  the  coefficient  of  the  radical 
part,  as  the  coefficient  of  the  other  factor  affected  by  the  given  radical 
sign, 

PROBLEM 


(212.)   1.  Reduce  4\/5(a' +a*6)  to  it  simplest  form. 

SOLUTION. 


Since,  V5(a'4-a*6)=V5(l  +a6)a'=Va'  V5(l  +a6)=al/6(l  +a6), 


we  have,  4V6(a''  +  a*6)=4aV5(l +a6). 

PROBLEM 

2.  Reduce  ^Vf  to  its  simplest  form. 

SOLUTION. 


Since,  Vf =VM=V2V  •  15=ViV  Vl5=iVl5,  we  ^*^®  W^- 


144-  REDUCTION  01*  StJRDS. 


EXAMPLES 


1,  Reduce  Vl6a^x  to  its  simplest  form.  Ans.  4:aVx, 


2.  Reduce  \/a¥x  to  its  simplest  form,  Ans,  hl/cLX. 

3«  Reduce  V81  to  its  simplest  form.  Ans.  3V3. 

4i  Reduce  1^288  to  its  simplest  form.  Ans,  12V2. 

5*  Reduce  V^  to  its  simplest  form.  Ans,  \VQ, 


6*  Reduce  V2^a^x^  to  its  simplest  form.  Ans,  3ax^V3ax, 


7.  Reduce  V5ax*—Sb'^x^  to  its  simplest  form. 


Ans.  xV5ax—3b^. 


8.  Reduce  Vo^'"'^^  to  its  simplest  form.  Ans,  aVa"&. 


9.  Reduce  V(a-\-bxyxi/  to  its  simplest  form.     Ans,  (a  +  bxy^xy. 


10«  Reduce  Via  +  xyb^  to  its  simplest  fonn.  Ans,  (a  +  x)b^. 


11.  Reduce  y- 7^—^ to  its  simplest  form. 

12.  Reduce  n/135  to  its  simplest  form.  Ans.  3\/5. 
13t  Reduce  51^64  to  its  simplest  form.  Ans,  15VQ. 

14.  Reduce  3v/108  to  its  simplest  form.  Ans,  9\/4. 

15.  Reduce  \/ax^-\-bx^  to  its  simplest  form.  Ans.  x\/a  +  bx\ 

16.  Reduce  ^V^  to  its  simplest  form.  Ans,  t^^I^21. 

17.  Reduce  6\/|  to  its  simplest  form.  Ans,  f  VIS. 

18.  Reduce  -ji/ -r-  to  its  simplest  fonn.  Ans.  -r^^d. 


19.  Reduce  y  j-, r  to  its  simplest  form. 

/  CLOl/  X          ————— 

20.  Reduce  4/ to  its  simplest  form.       Ans. Va(a—x). 

^  a—x                 ^  a—x      ^         ' 


ADDITION  OF  RADICALS.  145 


ADDITION    OF    RADICALS. 

•  -   »  > 
^  PROBLEM.      ' 

(21 3.)  To  add  radicals. 

RULE. 

Reduce  the  radicals  to  their  si?nplest''/6rm,  and  proceed   as  in 
addition. 

PROBLEM 

(214.)  1.  Add  together  V500  and  VTOS- 


SOLUTION. 


V500=V125  .  4=V125V4=:5V4 


and   V108~V27    ,  4=V2l  V4.=SV4. 
Therefore,       V600  +  \/T08  ==  5  V4  +  3  V4 = 81/4. 

PROBLEM 

2.  Find  the  sum  of  SVf  and  2V^. 

SOLUTION. 


and  2V^=2V^~=Wji^ro=2^rU^lO=2  .  ^yf^lQ:^  14/13 
Therefore,  3  V|  4-  2yjr=l^l^  +  if  10 =|f/10. 

EXAMPLES. 

1,  Find  the  sum  of  yTs  and  4/f.  Ans.  5V2, 

2.  Find  the  sum  of  |/T5  and  y48.  ^ws.  9|/3. 
3«  Find  the  sum  of  |/aV  and  4/c'ar.  Ans.  {ai-c)Vjs. 
4,  Find  the  sum  of  |/T50  and  -^64.  ^ws.  2 4/6. 


5.  Find  the  sum  of  Va'"^  and  V&a;"".  ^tw?.  (a  +  a:')V6. 

.  $,  Find  the  suna  of  j/iax'  and  3xV9a.  Ans^  llx^a, 

10 


146  ADDITION  OF  RADICALS. 


7.  Find  the  sum  of  3a;V2aV,  SaV^d'x^  and  2ax\/2a^x\     

Ans.  13a^V2aV. 

^    ^.    ,   ,  .     /a*x      /a^x^        ,      /a'c'^x 

8.  Find  the  sum  of  y  — ,  |/  --^,  and  y  -r^- 

la^      ax     ac\    /x 


».  Rud  the  sum  of  |/-,_^_-^    and  |/_--^-_^. 

An.  .g-±5).^. 

10.  Find  the  sum  of  V32  and  2 V40.  A'ns,  2V2  +  4V6. 

11.  Find  the  sum  of  ^^24,  \/bl,  and  — 1^6.  ^tw.  4|/6. 

12.  Find  the  sum  of  2VS,  -7fT8,  54^72,  and  -^^50.     ^«s.  8^2. 

13.  Find  the  sum  of  3V32  and  21/54.  ^«5.  6(^4"+ V2). 

14.  Find  the  sum  of  V24,  24/^2,  and  aV^\ 

Ans.  2{V^^-QVi)  +  axVQ, 

15.  Find  the  sum  of  8^1,  VOO,  —  Y'^^is'and  Vf        u4««.  44^3. 

16.  Find  the  sum  of  V81,  —2^24,'  ^28  and  2V63; 

Ans,  Sf'Y—VS. 

26  ^^   26 


17.  Fmd  the  sum  of  y  —-=-^  and  —y  — . 

Ans.  (3a- 1)  |/— . 

18.  Find  the  sum  of  Sb'VaFc,  +-  i^oV  and— c*|/^. 


19.  Find  the  sum  of  V54a'-»-*'6',  -Vl6a^-'b%  V2a*^%  and  V2c'a'". 

Ans.  {sa'b-—  +a'^+^+c\V2^. 


20.    Find     the    sum    of    V2'"flt'"^+'6'""+',    V3'"a"'''-^+''6'"+',    and 
-Va'6V».  ^W5.   (2a''6''-}-3a''-"6-c')Va^^"^. 


SUBTRACTION  OF  RADICALS.  147 

PROBLEM. 

(215.)  Find  the  sum  of  V^8  and  1^32. 

SOLUTION. 

If  we  square  V8  +  1/32  and  then  take  the  square  root,  we  would 
have  a  quantity  equal  to  the  sum  of  V8  and  1^32.  The  square  of  V8 
i-V32  is  8  +  2.V'8.V^32  +  32,  or  40  +  2  .  |/8.32,  or  40  +  2 1^256,  or 
40  +  2  .  16,  or  40  +  32  =  '72.  The  square  root  of  72,  or  V'72=:6f'2, 
which  is  the  sum  of  V8  and  f  32. 

The  application  of  this  mode  gives  exercise  in  the  involution  of 
radicals. 

EXAMPLES. 

1,  Find  the  sum  of  Vl2  and  V2l.  Ans.  Sf^S. 

2.  Find  the  sirni  of  V16  and  V54.  Ans.  5V2. 


SUBTRACTION    OF    RADICALS. 

PROBLEM. 

(216.)  To  subtract  radicals. 

RULE. 
Change  the  sign  of  the  radical  to  he  subtracted  and  proceed  as  in 
addition  of  radicals. 

PROBLEM. 

(2 1 T .)  From  V^  subtract  V}. 

SOLUTION. 

and       Vj  =  ¥~^\=^  V^\  .  6=}  V6  =  j\  V6 
Therefore,  V^—  Vi= j\  V6  —  j\  V6= j\  V6. 

EXAMPLES. 


1,  From  Vl08ax^  subtract  V48ax\  .  Ans.  2xVSa. 


14£(  MULTIPLICATION  OF  RADICALS. 

2.  From  9a Vb^  take  bxVc^b,  Am,  4axVh. 

3.  From  Va"6  take  Vb^-  Am,  (a— a;*)  V6. 


4.  Subtract  aVbc^  from  Vl6a*6V.  -4w5.  od^ft. 

5.  Subtract  3^45  from  61^20.  ^n«.  V5. 

6.  From  Vi92  subtract  1/24.  ^w«.  2V3. 

7.  From  SV'§  subtract  2V^.  Am,  f  VlO. 

8.  From  ||/|  subtract  |V^.  Am.  }^Ve, 


9,  From     y ——-  subtract  y  ttt-       Am.  l3ax—-\  y -^ 


ax 
2b' 


tA    17  /a'b  +  2ab^  +  b'     ^,  .A^— 

10.  From  a/ — 7  ^  ^     „  subtract  i/ —^- 

f      a''—2ab  +  b^  ^     a'  + 


a'b-2ab^+b^ 


2ab-\-b^ 

,       4cabVh 
Ans, 


a'-b'' 
11,  From  V66  subtract  — Vl89.  Am,  5  VI. 


12.  From  SVa'b  subtract  —SViea*b,  Am.  (12a»  +  3a)f'6. 


^  *»  ♦  >' 


MULTIPLICATION    OF    RADICALS. 

PROBLEM. 
(21 8.)  To  multiply  by  radicals. 

RULE. 

Reduce  the  radicals  to  equivalent  ones  expressing  the  same  root^  and 
multiply  the  coefficients  together  for  the  coefficient  of  the  product,  and 
the  parts  under  the  radicals  for  the  radical  part. 

PROBLEM. 

(2 1 9.)  Multiply  Va  by  V&. 

SOLUTION. 


V«*='"Va'*  and  V6=:'V6"',  whence  Var-V6='^ V«* ••"%/&-= "Va'*" 


MULTIPLICATION   OF   EADICAL8. 
EXAMPLES. 


149 


1.  Multiplyy2  by^/S. 

2.  Multiply  i/2  by  V2. 

3.  Multiply  |/2  by  V4. 

4.  Multiply  5  |/6  by  2  4/3. 

5.  Multiply  4/8  by  Vl6. 

6.  Multiply  2V|  by  3V|. 

7.  Multiply  4  4/3  by  3l/i. 

8.  Multiply  2  |/27  by  4/3. 

9.  Multiply  5a^  by  3aa. 

10.  Multiply  together  Va,  V*  and  Vc. 

11.  Multiply  together  V4,  ^Ve  and  ^1/5. 

12.  Multiply  together  4,  2V8  and  VY2. 

13.  Multiply  j^xhj  \/ax{a''—x^), 

14.  Multiply  together  y  4/aa;,  -^  V<^y  and  ^Vcz  . 

^7W.  xfVa'bVxyz\ 

15.  Multiply  a64-<J  4/a;y  by  a— 4/2. 

Ans.  a^b-{-ae  j^xy —ah/^z—c  j^xyz, 

16.  Multiply  re—  j/a^y+yby  |/a;+  |/y.  ^W5.  a;|/a:  +  y4/y. 
IT.Multiply  V^+«'  V6>  V«^V^+«&+  |/a  V6*+  W  by    4/a 


-4n«.  |/10. 

-47W.  V32. 

u4n5.  V128. 

Ans,    30  4/2. 

-4w«.  4V32. 

An9,  2V15. 

^n«.  12V432. 

^W5.  18. 

Am,  16^/a^ 

Ans,  |Vi20. 

-4»«.  8  j/6. 

Ans,  a'x  jJq^—q^, 


c'd.  — 


-Vb. 

18.  Multiply  4/2+  4/3  by  2  4/2-  4/3. 

19.  Multiply  2  4/6-3  4/5  by  4  4/3-  4/T0 


^W5.  1+  4/6. 

Ans,  39  4/2  — 16  4/I5. 
20.  Multiply  5  +  Vl  +  2 V5  by   4/6  +  4/5. 

^ns.  5  4/6  +  5  4/5  +  2Vl25+2Vi80-f  2V54  +  V2000. 


150  DIVISION  OF  RADICALS. 


DIVISIOlSr    OF    RADICALS. 

PROBLEM. 
(220.)  To  divide  one  radical  by  another. 

RULE. 

Reduce  the  radicals  to  equivalent  ones  expressing  the  same  root,  avd 
divide  the  coefficient  of  the  dividend  hy  the  coefficient  of  the  divisor  for 
the  coefficient  of  the  quotient,  and  the  radical  part  of  the  dividend  hy 
the  radical  part  of  the  divisor  for  the  radical  part  of  the  quotient, 

PR  OB  LE  M. 

(221.)  Divide  J  4/5  by  |V2. 

SOLUTION. 

i  |/5=iVi25,  and  |V2=|Vi.  Dividing  |  by  f,  we  obtain  f,  and 
Vi25  by  Vi,  we  get  V^^.  Therefore,  J  y5^|V2=|VH^ 
=tV-^6T^=|V2000. 

examples. 

1,  Divide  ^V5  by  %V2.  Ans,  im. 

2,  Divide  4V32  by  VI6.  Ans.  4Vl,  or  2V2. 

3,  Divide  6^12  by  1/24.  Ans,  6V3. 


4.  Divide  4Vax  by  Wxy,  Ans.  - — Va^xy, 

5,  Divide  ^ax^  by  j^bx,  Ans.  Wab, 

0 


6.  Divide  \/ax—x''  by  Va^—x\  Ans. 


VxVO'—x 
%/(i->tx) 


7.  Divide  a/^-^^  by  6|/?^.  ^^.  V^fll^. 

^        c        ^     y        d  ar   hc{d—x) 

8.  Divide  a^x  —Vhx  +  af^y  — V%  by  Vx  +Vy.  Ans.  a  —Vb, 


INVOLUTION   OF   KADICALS.  151 

9.  Divide  a  +  b—c  +  2Vab  by  fa  -[-Vh  —Vc.      Ans.  4/a  +V6  -f  Vc. 

10.  Divide  34/15-1^20+1/10-7  by  2Vb. 

Ans.  ?f 3- 1+1^2-^^^246. 

1 1 ,  Divide  ^8  + 1/12.+  V2  by  W2.  Ans,  1  +  ^ VTs  +  ^ VB. 


2 


12.  Divide  ^/a"— ar'  by  a— a;.  -4?is.  4/- 


+2- 


a — X 


13.  Divide  ^ab^—b^c  by  4/a— c.  ^»s.  5. 

14.  Divide  i|/I  by  >i/2  +  3fi.  -^r^s.  yV- 

15.  Divide  4/72 +i/32— 4  by  |/8.  ^**s.  5— 1/2. 

16.  Divide  4^12  by  2^3.  Ans.  2 VY- 

n.  Divide  |/^  by  |/|.  ^m  j/^. 

18.  Divide  V64  by  2.  ^?^s.  \/2. 

19.  Divide  a  by  \/a.  Ans.  ^a. 


20.  Divide  %/ah^'c'  by  |/ ^-^^i-  -^»*»-  V i — . 


INVOLUTION"  OF    RADICALS. 

PROBLEM. 
(222.)  To  raise  a  radical  expression  to  a  given  power. 

RULE. 

Proceed  according  to  the  method  given  in  Involution^  observing  the 
rules  just  given  in  reference  to  radicals. 

PROBLEM. 

(223.)  1.  Square  SVB. 


162  INVOLUTION  OF  RADICALS. 

•     ' SOLUTION. 

It  is  evident  that  the  square  of  any  quantity  is  equal  to  the  product 
of  the  square  of  its  feclors.  If,  then,  we  multiply  the  square  of  VS  by 
9,  we  must  have  the  desired  result.  We  know  from  the  nature  of 
fractional  exponents,  that  the  square  of  the  cube  root  of  a  quantity  is 
eqtial  to  tlie  cube  root  of  its  square ;  or,  in  algebraic  language, 
(Vcf)-  =  \/a',  because  {aiY=ah  Hence,  the  square  of  V3=V9  and 
consequehUy  (3V3)'  =  9l/9. 

/  PROBLEM 

.  2,  Raise  |/3— 1^2  to  the  4th  power. 

-     ,  SOLUTION. 

By  the  Binomial  Theorem,  we  have 

{v3-V2y={v~3y-4{V3y{v~2)  +fi{vsy{V2y^4:(Vs)(V2y + {V2y.   ^ 

Sin^lifying  this  by  the  rules  for  radicals,  we  obtain 

{VS—V2y=9—12y6-\-SQ  —  SV6  +  4t=49  —  20V6, 

EXAMPLES. 

'i.  Raise  }V6  to  the  4th  power.  Ans,  ^. 

.2.  Raise  —v/a=' to  the  4th  power.  Ans.  a^X/a^. 

3.  Raise  1*71/21  to  the  3d  power.  Ans.  1031731/21. 

4,  Cube  j^x + dVy.  Ans.  xVx  +  2lyVx  +  QxVy  +  2 1yVy> 

5t  Cube  —A/Va—Vhc.  Ans.  \/bc—Va. 

6.  Square  Vaa;*.  Ans.  j^ax*. 

7.  Cubei^i  Ans.  j/(«+^)(^^> 

VaVa+x  «' 

ah 


8.  Raise  — ^  V{c  +  xy  to  th6  4th  power. 

X 

Ans.  — g-(^+^)*» 

X 

9.  Cube  aj/ft"— a;'  +  3a;V««'.     '•     Ans.  a'6«— aV  +  3a'a:Va?. 
10.  Square  |/6+V3.  Ans.  S^WU. 


EVOLUTION  OF  RADICALS.  158 


EVOLUTION    OF    RADICALS. 

PROBLEM. 
(224.)  Extract  the  mth  root  of  a  given  monomial  radical. 

RULE. 

Extract  the  mth  root  of  its  coefficient,  atid  multiply  the  result  hy 
the  mtli  root  of  the  radical,  which  is  obtained  hy  multiplying  the  index 
of  the  radical  hy  the  index  of  the  root  to  he  extracted,  leaving  the 
quantity  under  the  radical  sign  unchanged  ;  or  hy  extracting  the  mth 
root  of  the  quantity  under  the  radical  sign,  leaving  the  radical  sign 
unchanged. 

PROBLEM. 

(225.)  To  extract  the  mth  root  of  6V«^ 

SOLUTION. 

m.    / rr         \  1  "  /~7 

Following  the  rule,  we  get  i/6v/a'=6'»'"Va'— S«y  ««.  When 
r=m,  we  have  juhVa'^=.h'^  Va, 

EXAMPLES. 

1,  Extract  the  square  root  of  |/a.  Ans,  \/a. 

2.  Extract  the  cube  root  of  |/a\  Ans.  Va, 

3*  Extract  tlie  square  root  of  ym'n,  Ans.  i/rnVn. 

4*  Extract  the  square  root  of  |/a*6'.  An^.  af/\ 

5*  Extract  the  square  root  of  %\Va.  Ans,^^\/a. 

6.  Extract  the  square  root  of  91/3.  Ans.  3V3. 

7.  Extract  the  4th  root  of  jf Vo^  Ans.  \Va. 

8.  Extract  the  cube  root  of  (5a''  — 3a;'*) 2.  Ans.  ^/Ea'^—Sx''. 

9.  Extract  the  cub6  root  of  i«'  Vb.  Ans.  ^aVh. 
10.  Extract  the  4th  root  of  iea"  Vx}  Ans.  Wa^^ 


164                               EVOLUTION   OF  EADICALS. 

11.  Extract  the  5th  root  of  32V^  Ans,  2VS. 

12.  Extract  the  nth.  root  of  VaV.  Ans.  am  x  mn, 

13.  Extract  the  cube  root  of  -y  -.  Ans.  ^Sa. 

3       S 

14.  Extract  the  cube  root  of  i  1/2.  Ans.  iV2. 


15.  Extract  the  cube  root  of  {a-i-x)Va+x,  Ans.  \/a  +  x. 

16.  Extract  the  cube  root  of  "V8^'.  Ans,  ^V^. 


17.  Extract  the  cube  root  of  27  V27a».  Ans.  3  V3a'. 

18.  Extract  the  4th  root  of  81a"  V4?7  Ans.  3a''  \nx. 


19.  Extract  the  18th  root  of  Va'°6".  ^tw?.  a  '\/h\ 


20.  Extract  the  5th  root  of  4/320;'".  ^rw.  a;v/2. 

PKOBLEM. 

(226.)  To  extract  the  square  root  of  a  binomial  surd,  one  of 
whose  terms  is  rational  and  the  other  a  quadratic  surd. 

SOLUTION . 

Let  \/x-^Vy  represent  the  square  root  oi  a-^Vh. 
Then,  x  +  2Vxy  +  y,  or  a; 4-2/  +  ^Vxy^^a  +  ^h ;  whence,  by  Theorem, 
(204),  we  have  x-\-y^=.a.     We  know,  by  Theorem  (205),  that 

if  |/a;  +  f/y = i/a  +4/6,  that  /^x  —Vy= ju  a — 1/6.     It  is  evident,  then, 

that  the  product  of  j^x^Vy  by  i^x—Vy^  which  is  x—y^  must  equal 
the  product  of  A/a+Vb  by  A/a—Vb,  which  is  ^a^—b.     This  gives 

us  a;— y=V'a'^  — 5.     If  we  extract  the  square  root  of  half  the  sum  of 
x-^y  and  x—y,  the  result  must  equal  the  square  root  of  half  the  sum 

of  a  and  >^V— 6,  that  is  ^x=y . 

Also,  if  we  extract  the  square  root  of  half  the  difference  of  x-\-y 
and  x—y^  the  result  pust  equal  the  square  root  of  half  the  difference 


<Qf  a  and  j^a^^h,  that  is,  A/y=4/  - 


-Va'-b 


EVOLUTION   OF  RADICALS.  155 

Whence,  we  have  the  formula, 


Letting  4/ a—Vb=zVx  —Vj/,  we  would  have 


.^/;z:k^-/^-h^«^-^  /^-^^^ 


■b 


2 

PROBLEM 

(227.)  1.  Extract  the  squai-e  root  of  19 +  81^3. 

SOLUTION. 

In  this  problem,  a=l9,  and  Vb=z8VS=Vl92,  or  6=192.     Putting 
these  values  in  the  first  of  the  above  formulas,  we  have 


^I^7^;T^^/19+4^-192^^19^-^ 


Instead  of  using  the  formula,  we  might  pursue  the  operation  indi- 
cated in  obtaining  the  formula. 

PROBLEM 

2.  Extract  the  square  root  of  7  +  4V3. 

SOLUTION. 

This  may  be  solved  as  the  last  problem.  The  following  method, 
however,  is  generally  as  easily  applied  : 

7  +  41^3=4  4-41/3  +  3.  Taking  the  square  root  of  4  +  41^3  +  3 
by  the  rule  given  in  Evolution,  we  have  for  the  result  2  +  VS. 

It  is.  easier  to  use  the  fact  that  a  trinomial  is  a  perfect  square  when 
twice  the  product  of  the  square  roots  of  two  of  the  terms  is  equal  to  the 
remaining  term. 

lu  this  example  a  bare  inspection  shows  that  twice  the  product  of 
the  square  roots  of  4  and  3  is  equal  to  41^3. 

Hence,  i/'7+4V'3,  or  a/4+4VS-\-S=:2+VS. 


^^ 


156  EVOLUTION  OF  RADICALS. 


^_ PJIOBLEM 

8.  Extract  tW  square  root  of  V'32—V24'. 

SOLUTION. 

It  is  obviggs  that  the  formula  can  not  be  applied  to  /^32— f^24  as  it 
stands,  since  both  terms"  ^re  surds.  It"  niay^  however,  be  made  to 
assume  the  proper  form,  for  " 

4/32  — 1/24=21/8— V3V8  =  |/8(2  — 1/3). 

If  we  extract  the  square  root  of  2—V3,  which  is  of  the  requisite 
form,  we  have,  by  the  second"  formula,"   ' 

|/2-|/3=i/|-./I=Vf-Vi. 

.    -This,  multiplied  by  the  square  root  of, 4/8,  which  is  V8,  gives 

Vi8-V2. 

Therefore,  yV32-y24=Vl8-V2. 

PROBLEM 


4.  Extract  the  square  root  of  a^  +  2xVa^—a:^. 

S  OLUTION. 


Since   a^  +  2xVa^ —x'=x\t\-2xV(i'  —x' +d'—r'y  a  bare  inspection 
shows  that  its  square  root  is  yi-^ya^—x'^, 

EXAMPLES. 

1 ,  Extract  the  square  ;cpot  of  6  +4/20.  .      .        4ns.  1  +f  5 . 

2,  Extract  the  square  root  of  8  +|/39.  Ans.  1(1/26  +f  6). 

3,  Extract  the  square  root  of  11  4-4/72.  Ans.  3  +V2, 
.4.  Extract  the  square  root  of  6  —  2 V'5.  Ans.  —14-1/5. 

5.  Extract  the  square  root  of  23  — 8I/7.  Ans.  4— VI. 

6.  Extract  the  square  root  of  33  4- 124^6.  Ans.  3  4-  24/6. 

7.  Extract  the  square  root  of  1  —  2V10.  Ans.  \/b—V2. 


8,  Extract  the  square  root  of  42  4- 34/1741.        Ans.  |/14  4-24/7. 

9.  Extract  the  square  root  of  10— V^.  Ans.  —2  4-4^6. 


iEVOLUTION  OF  RADICALS. 


f^ 


Extract  the  square  root  of  x^^^Vx—t. 
Extract  the  square  root  of  28  +  5^12. 


to. 

tl. 

12.  Extract  the  square  root  of  2  +  2(1  —a;)  1^1  +2a;— .r^ 


Ans,  1— -Vi— 1. 

Ans.  5  4-V3. 


13. 

15. 

16. 
*  17. 
'  18. 

19. 

20. 

21. 

22. 

23. 

21. 

25. 


Ans.l—x+Vl  +  2x-rps\ 

Extract  the  square  root.pf  43  — 151/8.  Ans.  6—^V2. 

''''.  '■''  J  u  ___  _        _ 

Extract  the  square  root  of  5 —1^24.  .      ,.Anf»  V3-ry2, 

Extract  the  square  root  of  3— 2^/2.  '^  Ans,  'lr^2» 

Extract  iflie  spare  ,ropt  of  87  — 12^^42^^  Ans.  SV'7-r-2VS, 

ExtraV!t  the  square  root  of  f  +V2J '  Ans.  1  -{-'^■11^2. 

Extract  the  square  root  of  2  +VS.  Ans,  ^V6-\-^V2, 

Extract  the  square  root  of  4/27  +  2 V6.  Ans.  Vl 2  +  i/S. 

Extract  the  square  root  of  |/32  +  6.  Ans.  2  -\-V2, 

Extract  the  square  root  of  3V^54- 1^40.  Ans.  \/5  +  V20. 

Extract  the  square  roof  6f '3 V6^-^' 2  Vl2.  An^.  V6  +  V24. 

Extract  the  square  root  of  4/1 8— 4.  Ans.  V8  — V2. 

Extract  the  square  root  of  12—1^140.  Ans.  |/7— V5. 


Extract  the  square  root  of  2a  +  2Va^—b^ 


Ans.    j^a-\-b  -^Va — b. 


26.  Extract  the  square  root  of  (ix  —  2aVax—a^ 


Ans.  a^-Vax- 


27. 


Extract  the  square  root  of  "t  +  o^^'"^'- 


Ans.  -  +  ±Va'-(j'. 


28.  Extract  the  square  root  of  (xi-ii;i/)  —  2xVi/. 


Ans.  {yy—l)Vx, 


0„  I 

29.  Extract  the  square  root  of  -^  +  /i/ 


12aV_4aV 


ac        /3a     a'c* 


158  EVOLUTION   OF  RADICALS. 

30.  Extract  the  square  root  of  1 6' — a6  +  -^  j  4-^4a6' — 8a'6'  +  a%. 


r 


Ans.  j^/ah  ■\-Vb^ — 2ab + J-a'. 

PROBLEM. 
(228.)  To  extract  the  cube  root  of  a  binomial  A  +  B, 

RULE.* 

"  Separate  either  term  as  A^  into  two  such  parts  that  the  one  of 
them  may  be  a  cubic  number,  and  the  other  part  divisible  by  3  with- 
out a  remainder;  then  the  cube  root  of  the  said  cubic  part  will  be  one 
term  of  the  root,  and  the  other  term  will  be  the  square  root  of  the  quo- 
tient, arising  from  dividing  the  aforesaid  third  part  by  the  first  term 
just  found.     So    if   A   be  divided  into  r^4-3s   then  the   root  is 

PROBLEM . 

(229.)  Extract  the  cube  root  of  10 -f-  Vl08. 

SOLUTION. 

Separate  10  into  the  two  parts,  1  and  9  of  which  the  first  is  a  per- 
fect cube,  and  the  other  exactly  divisible  by  3  ;  whence  r=l  and  5=3. 

Therefore  r-f-|/  *=1  +1^3.  We  can  obtain  the  same  result  by  separa- 
ting f  108  into  two  parts,  f  108=  V2l  +  3f3,  the  first  of  which  is 
a  perfect  cube,  and  the  second  exactly  divisible  by  3  ;  whence,  r=  4/3 

and  s=  4/3.     Therefore  r  +  r  -=  4/3  +  r  -==f3  +  l. 

EXAMPLES. 

1.  Extract  the  cube  root  of  26  + 15  |/3.  Ans.  2  +  -/s. 

2.  Extract  the  cube  root  of  ^  |/3  — 11  4/2.  ^ns.  V3— V2. 

3.  Extract  the  cube  root  of  135±78|/3.  Ans.  3^=^12. 

4.  Extract  the  cube  root  of  72  +  1/5120.  Ans.  3  +  V'5. 

*  This  rule  is  given  by  Hutton  in  his  Mathematical  Dictionary,  and  is  credited 
to  Tortalea. 


EVOLUTION  OF  RADICALS.  159 

1  +  1/5 


5,  Extract  the  cube  root  of  8  +  4  V5,  Ans. 


V2 

4:-V6 


V2 


6.  Extract  the  cube  root  of  68  — 1^43 74.  Ans. 

PROBLEM. 
(230«)  To  extract  any  root  (c)  of  a  binomial  surd. 

RULE.. 

^Let  the  quantity/  be  A±:B,  whereof  A  is  the  greater  part^  and  c 
the  expcment  [index]  of  the  root  required.  Seek  the  least  number  in 
whose  power   n"  is  divisible  by   AA—BB  \A'^—B'^\   the    quotient 

being  Q.     Compute  J/A  +  BxVQin  the  nearest  integer  number, 

which  suppose  to  be  r.     Divide  AVQ  by  its  greatest  rational  divisor, 

r  +  - 
and  let  the  quotient  be  s,  and  let  -,  in  the  nearest  integer  num- 


be7',  be  t ;  so  shall  the  root  required  be ^^: if  tJie  c  root  of 

ve 

A:^B  can  be  extracted.'''' 

Remark. — This  rule  is  taken  fronr  Newton^s  Algebra,  p.  59.  It 
is  there  given  without  any  demonstration.  Maclaurin  has  attempted 
a  demonstration  of  it  in  his  Algebra^  p.  120  ;  Mr.  Ryan  says  the  rule 
"feils  when  t=i\  exactly;  in  which  case  instead  of   taking  t  the 

r-\-  — 
nearest  integer  value  of -,  it  must  be  taken  equal  to   ^."     He 

says  he  proposed  to  the  New  York  Mathematical  Club  the  following : 
"  Required  to  know  if  the  cube  root  of  2  4^7  +  3  VZ,  can  be  found  by 
the  rule  given  by  Newton,  p.  59,  Universal  Arithmetic,  for  extracting 
any  root  of  a  binomial  surd ;  and,  if  not,  to  show  where  that  rule 
fails,  and  what  alteration  is  to  be  made  in  it,  so  as  to  obtain  the  root  ?" 
Mr.  Ryan  states  that  Dr.  Adrian  "  ably  investigated  the  subject,  and 
found  the  rule  not  only  to  fail  in  this,  but  in  a  great  variety  of 
other  examples ;  and  also  discovered  the  rule  to  be  defective." 

PROBLEM 

(23 1 .)   1.  Extract  the  cube  root  of  1^^968  +  25. 


^160  .livoxtmoN^  OF  RA-mcAfiS. 

'"  tJ^ -!    I  •  SOLUTION.  " 


r  \( 


Here  (f/968)'  — 25'=^'— -6*=343.  It  is  evident  since  the  divisors 
^f;;3.43  aje  7,  7,  T  ;  that  7 Ms  the  least  number  of  the  form  n^  that  is 
exactly  divisible  by  343,  whe©<?o.^^\v:e  have  .w=;:7,  au^  §=1. 
{A  +  B)VQiov  V^968  +  25  is  a  very  little  more  than  66,  of  which  the 
nearest  cube  root  is  4,  tjierefore .  i^p?4.  Dividing  V'968=22V^2 
=AVQ  by  itsrgreatest  ratioi^al  divisor  22,  we  obtain  1^2 =»,  whence 


'  ^  »•  or  — — 
2s  2V2 


in    the   nearest    integer     is    2  =  ^.     Then   te=2f^2, 


V^V-?I=1,  and  V§=Vl  =  l.     Hence  2*/2  +  l,  or  l+2f2  is  the 
cube  root  of  1^068+25,  which  result  will  be  proved  by  trial. 

PROBLEM 

2.  Extract. the  fifth  root  of  29  f^6 +41  ^3. 

SOLUTION. 

Here   .4*-^'=3,   and  w=3 ;    §=81;    r=5 ;    8=VQ\    «=1  ; 
^t8=  V6  ;  fW— 9i=f  3  ;  and  'x7§= V81= 1/9.     Consequently   trial 

must  be  made  with  ^^i±^,  or  ^l±i:!. 

,-....  EXAMPLES. 


1.  Extract-th^^ct[berootofy242  — 12f  Ans.  ^^^     ^ 


V2 

2.  Extract  the  cube  root  of  11  +6^7. .  Ans.  -Jliili*. 

3.  Extract  the  fourth  root  of  49849—28^5  |/224. . 

Am,  5V1SV2, 

4.  Extract  the  cube  root  of  21^7+31/3.  Ans,  ^   '^     . 

•.  •■>  ot  m  o?  ,li  !?i  0  ■-':?r  wl  <  j  .        2  .     - 

■  ■ .  •    ■ '  '-'  ^'--y    'hnc)ltE^M\  •  ■  •  '  :.^:-  ;„/.\ 

(432.)  To  nnd'sucK  a  multiplier,  orsucHinuIilpBere  ^  will  ifilake 

any  binomiarslird  rational.  .  -  o-n  i  I'o -^v'.o 

All  binonaial  siifds,  not  imaginary,  may  be  represebteS  ty^'^aitvi. 


EVOLUTIOJSr  OF  EADICALS.  161 

Let  us  then  seek  a  general  expression  for  a  multiplier  which  will  ren- 
der "y/aztiy/h  a  rational  quantity.     We  know  by  Theorem  5,  (1 1 3), 

r  r 

dm —  On  .  -,  . 

that =V>   ^^  exact   quotient  when   r  is   an  even  number. 

am-\-  bn 

1  \.  L  L 

§  is  a  multiplier  which  will  make  a^-^bn  equal  to  a»«— 6».    But  we 

r  r 

wish  am—b^  to  be  a  rational  quantity,  which  it  will  be  when  r  is  a 
multiple  of  both  m  and  n.  In  this  case,  however,  we  must  take  for 
r  a  multiple  of  m  and  n,  that  is  also  an  even  number. 

r  r 

By  Theorem  6,  (114),  we  know  that  — -=Qi  an  exact 

quotient  when  r  is  an  odd  number.  If  r  is  taken,  an  odd  number, 
such  that  it  is  a  multiple  of  both  m  and  w,  §,  is  the  multiple 

1.        JL 
necessary  to  render  am-{-bm  rational. 

r  r 

We  know  by  Theorem  4,  (112),  that  -^ ^=§2  ^  ®^^* 

dm —  On 

quotient,  r  being  any  positive  integer.    If  then  r  be  taken,  any  multi- 

i_       i_ 
pie  of  both  m  and  n  Q^  is  the  multiplier  necessary  to  render  a«— 6i». 

In  each  of  these  cases  it  is  evident  that  r  may  be  assumed  to  be 

any  of  the  indefinite  number  of  values  which  can  be  found  to  fidfill 

the  required  conditions.    The  least  of  these  values,  however,  is  the 

one  generally  used. 

PROBLEM 

(233.)  1.  Find  a  multiplier  that  will  render  6+1^7  rational. 

SOLUTION. 

A   simple  inspection  shows  that   the    multiplier  is   6— VI,  for 
(6+V1){6—V1)=36  —  1=29,  a  rational  quantity. 

r  r  r  r  r 

Afifain,  we  have = = ,  rmust  be  assumed  to 

am-if-b'^     61  +  7^      6 +  '7* 
be  some  even  number  which  is  a  multiple  of  1  and  2.    Making  r=2, 

we  have -=6— T^  =  Q  as  before. 

6  +  V^ 

11 


162  EVOLUTION  OF  RADICALS. 

PROBLEM 

2.  Find  a  multiplier  that  will  render  2V3— 3V2  rational. 

SOLUTION. 

r  r  r  r 

alk—lTn      122—543 


Here  2^/3 — 31/2  =1^1 2 — V54 ;  hence,  we  have 


am— 611     122—543 


122—543        12^— 54^^ 
= =z .     Applying  the  formula  given  in  the  de- 

122—543      122-543 

12^-54'  5 

mostration    of   Theorem    4,    (112),    we    get    — j ^=1224- 

122—543 

122543  +  122543  -f  122543  +  122543  4-  54^  =  V12*  +  144V54  + 
V'T2'V5?  +  648+Vi2V54'  +  V54^  which  simplified  is  28»^'3  + 
432V2  + 2161/3^4 +  648  +  324V3V2  +  486V4. 


EXAMPLES. 

1,  Find  a  multiplier  that  will  render  Vh-\V^  rational. 

Ans.  |/5— f^3. 

2.  Find  a  multiplier  that  will  render  5+1^3  rational. 

Ans.  5— V3. 

3«  Find  a  multiplier  that  will  render  V2—Vx  rational. 

Ans.  ^2+Vx. 

it  Find  a  multiplier  that  will  render  aVh  +  hVa  rational. 

Ans.  aVh—hVa. 

5«  Find  a  multiplier  that  will  render  1— V2a  rational. 

A71S,  l+V2a  +  V4a^ 

6.  Find  a  multiplier  that  will  render  V3— ^1/2  rational. 

Ans.  V9  +  iV6  +  iV4. 

7.  Find  a  multiplier  that  will  render  V5  +  \/3  rational. 

Ans.  (V5-V3)(t/5+t/3). 
8i  Find  a  multiplier  that  will  render  VoN- V^rational. 

Ans.  V'cf-V'a'^+V^'-V^- 


EVOLUTION   OF  RADICALS.  163 

PROBLEM. 

(23  4t)  To  reduce  a  fraction  whose  denominator  is  a  surd  quan- 
tity to  another  that  shall  have  a  rational  denominator. 

SOLUTION. 
A  simple  fraction  which  has  a  monominal  surd  for  its  denominator, 
may  be  represented  by  — =,  in  which  a  may  represent  any  quantity 
whatever. 

Now,   if  we   multiply   both    terms   of  —^   by   V^"~S   we  have 

Vx 

=zr-=: ,  which  is  a  fraction  having  a  rational  denominator, 

if  X  is  rational.     If  ic  is  a  binomial  surd,  it  must  be  rendered  rational 
as  in  the  last  problem, 

PROBLEM 

2 
(235,)  1.  Reduce  — r  to  a  fraction  having  a  rational  denominator. 


SOLUT ION. 

—  2y'3 

Multiply  both  terms  by  |/3,  and  we  have . 

3 

PROBLEM 


2.  Reduce  —         -  to  a  fraction  having  a  rational  denominator. 

24/2  +  ^/3- 

8OL  UTION. 


1^8+^48     (8  — f48)i/8+V48 


21/2  +  1/3      f8+V'48        8+V/48  64-48 

4(2-1/3)21/2 +|^3^(2-|/3)|^2+l/3_  ,^-— ;^ 
16  ~  2  ~^ 


XAMPLES 


It  Reduce  -^—^  to  a  fraction  having  a  rational  denominator. 

Ans.  |V125. 


164  .^n^EVOLUTION  OF  RADICALS. 

3 

2,  Reduce  -^r =.  to  a  fraction  having  a  rational  denominator. 

Ans,  4/5  +  ^2. 

3*  .Reduce  — — --  to  a  fraction  having  a  rational  denominator, 
B-V2 

.       2  +  3^2 

Ans.  — . 

7 

Q 

4.  Reduce  -^ to  a  fraction  having  a  rational  denominator. 

V5-V2  _ 

Ans.  V26  +  \/io  +  V4. 

5,  Reduce  ^^+^  --Va—x  ^  ^  fraction  having  a  rational  do- 

j^a+x  +Va—x  

.       a—Va'-^x^ 
nominator.  Ans.  • 


6«  Reduce  y^  +x+l  +yx     x       ^^  ^  fraction  having  a  rational 

x^-ir^x'^—x^-'lx—X 
denommator.  Ans.  — • 

2 

7,  Reduce  — = = =  to  a  fraction  having  a  rational  denom- 

4/6  +  1^3-^2  _  _       _ 

.       2V3-3t/2  +  V30 
mator.  An%.  — •• 

8#  Reduce  -^ to  a  fraction  having  a  rational  denominator. 

V3-V2 

Ans.  V9  +  V6+V4. 

8 
9,  Reduce  — = = to  a  fraction  having  a  rational  denomina- 

4/3  +  ^2  +  1 
tor.  AnB.  4  +  2y2-2f^6. 

10.  Reduce  — =: to  a  fraction  having  a  rational  denominator. 

V6+V3  _        _       _ 

Ans.  V126-V'76+V45-V2Y. 

PROBLEM. 

(236.)  Transform  2—^3" to  a  general  surd. 

SOLUTION. 

Squaring  2  — f3,  we   have  7 —41/3;  if  now  we   indicate  the  ex- 


IMAGINARY  QUANTITIES.  165 

traction  of  the  square  root  of  this  quantity,  we  shall  have  2  —  VS 
expressed  in  the  form  of  a  general  surd,  that  is  2—  VB=y1—4:VS, 

EXAMPLES. 

1.  Transform   |/a— 2  Va;  to  a  universal  surd. 


Ans,  y  a  +  4x—4:Vax, 


I 


2.  Transform  SVi+V1^  to  a  general  surd.  Ans,  SVS. 

3*  Transform   |/2 7  +  1/48  to  a  universal  surd.  Ans.  ^VS, 

4.  Transform  V320— \/40  to  a  general  surd.  Ans.  21/6. 

5«  Transform  |/2  +  VS  to  a  general  surd. 

Ans.  1/5  + 2  V6. 

6«  Transform  v/2  +  \/4  to  a  general  surd. 


^7is.|/6(l+V2+V4). 
7.  Reduce   4/2— 21^6  to  a  general  surd.        Ans.  y2e—SV8. 
J;,  j8.  Reduce  4—^7  to  a  general  surd.  Ans.  A/ 23— SVI. 


9*  Reduce  j^2+V3  to  a  general  surd.  Ans.  |/ 5+2i/6. 

10.  Reduce  2V3— 3v/9  to  a  general  surd. 

Ans.  |/l62V9-108V3-219. 


^  '«  ♦  •' 


IMAGINARY   aUANTITIES 

ADDITION  OF  IMAGINARY  QUANTITIES. 

PROBLEM. 


(237.)  To  add  y-a'  and  ^-b^  together. 

SOLUTION. 


Since,     i/-a'=Va\-l==aV-l,  and  ^-b'=Vb\-l=hV-ly 
we  have  |/^''  -f  V'lIb'—aV—l.  -^iV^l  =z{a-\-b)V—i. 


166  SUBTRACTION  OF  IMAGINABY  QUANTITIES. 

EXAMPLES.       , 

1.  Find  the  sum  of  |/— 4  and  4/— 9.  Ans.  ol^— 1. 


2.  Find  the  sum  of  2  +  f/-l  and  S  —  V—M.    Ans.  5  — IV— I. 

3.  Find  the  sum  of  ^—^  and  4/— 18.  Ans.  5^^. 


4.  Find  the  sum  of  ^f — 27  and  2V—\2.  Ans.  ISf— 3. 

5.  Find  the  sum  of  4/~6  and  4/^9.  ^?is.  (3  +  f  6)  V^. 

6.  Find  the  sum  of  4/ -5  and  4/~7.         ^»wf.  (1^5+  f' 7)  l/"^. 

7.  Find  the  sum  of  V— 4  and  V— 9.         ^n«.  (t/2+  y3)V^. 

8.  Find  the  sum  of  a  -\-V—h  and  a-^V—c. 

Ans.  2a-\-(Vb-\-Vcy'^. 

9.  Find  the  sum  of  V— a  and  2V— «.  Ans.  SV—a. 


10.  Find  the  sum  of  V-1  and  V-16.  Ans.  3V— 1. 


^  '«  ♦  »«  » 


SUBTRACTION   OF  IMAGINARY  QUANTITIES. 

PROBLEM. 

(238.)  From  4/— a"  subtract  4/"^'. 

SOLUTION. 

Since,     4/— a'^^a  4/— 1,  and  4/— 6'=6  4/— 1,   we  have  aV-^h 
—6  4/^=:(a— 6)  4/— 1. 

EXAMPLES. 

1.  From  4/— 9  subtract  4/— 4.  ^ns.  i^— 1. 

2.  From  3— V— 64  subtract  2 +V^.  ^»*«.  1  —  94/1. 

3.  From  4/— 18  subtract  4/— 8.  ^w«.  4/— 2. 

4.  From  4|/~27  subtract  24/— 12.  ^rw.  8V^. 

5.  From  a  +  V^  subtract  a  +  4^^  ^ws.  (f^6  — V^c) V—  1. 

6.  From  V-16  subtract  V^.  -4ws.  V-^i. 


MULTIPLICATION  OF  IMAGINARY  QUANTITIES.  16? 


MULTIPLICATION    OF    IMAGINARY 
QUANTITIES. 

THEOREM. 

(239.)  An  imaginary  expression  of  the  form  ^—A  can  always  be 
reduced  to  the  form  r^  —\. 

DEMONSTRATION. 

Let  r  denote  the  square  root  of  ^-A.  It  is  evident  that  r  can 
always  be  obtained  either  exactly  or  approximately  when  ^  is  a  posi- 
tive rational  quantity.     Then, 


^^A=V  +  A.  -l=VAV—l=rV—l, 

PROBLEM. 

(240.)  To  ascertain  the  rule  of  signs  in  the  multiplication  of 
imaginary  quantities. 

SOL  UTION. 

Let  |/— a  be  multiplied  by  y— a.  Multiplying,  as  in  the  case  of 
surds,  not  imaginary,  we  have  |/— a  xV—a=V—a  x  —a=Va^=—a. 
We  put  the  square  root  —a,  because  the  d^  was  obtained  in  this  case 
by  multiplying  —a  by  —a,  or,  what  is  the  same  thing,  by  squaring 
—a.  It  may  also  be  observed,  that  |/ — a  x  V — a  =:{V— ay.  We  have 
already  shown,  in  Evolution,  that  (V^— a)^=— a,  which  result  agrees 
with  that  just  given.  Let  us  now  ascertain  the  product  oi  \/—a  by 
V—h.  Since,  \/—a^=VaV—\^  and  \/—h=^VhV—'\.^  we  have 
V'^ xV'^=VaV'^  xVlV'^=V^V^-iVaVb==  —  lVab=-4^ab, 

Whence,  we  see  that  the  same  rule  applies  in  multiplying  quadratic 
imaginary  surds  as  in  other  surds,  provided,  however,  that  the  result 
must  be  affected  by  a  minus  sign.  The  principles  already  developed 
will  enable  the  student  to  multiply  any  imaginary  quantities  in  which 
the  index  of  the  root  is  even. 

In  all  these  operations  great  care  must  be  taken  that  all  the  steps 
be  rigid. 


168         MtJLTIPLICATIOK  OF  IMAQINABY  QUANTITIES, 

Note. — The  importance  of  carefully  scrutinizing  all  the  operations  in 
which  imaginary  quantities  are  concerned  can  not  be  better  set  forth  than  by 
showing  the  positions  assumed  by  different  distinguished  mathematicians : 

"  The  first  idea  that  occurs  on  the  present  subject  is,  that  the  square  of 
^—3,  for  example,  or  the  product  of  v/— 3  by  >/— 3,  is  —3;  and,  in 
general,  that  by  multiplying  ^—a  hy  y/—a,  or,  by  taking  the  square  of 
y/—a^  we  obtain  —a.  *  *  *  *  *  * 

"  Moreover,  as  y/a  multiphed  by  ^yh  makes  ^aJ,  we  shall  have  y/Q  for 
the  value  of  ^Z— 2  multiplied  by  v/— 3 ;  and'-v/4,  or  2,  for  the  value  of  the 
product  ofy^— 1  by  ^—4.  Thus,  we  see  that  two  imaginary  numbers, 
multiplied  together,  produce  a  real,  or  possible  one. 

"  But,  on  the  contrary,  a  possible  number,  multiplied  by  an  impossible 
number,  gives  an  imaginary  product;  thus,  ^—3  by  v'  +  S,  gives  y/ — 15." 
— Enter's  Al,  p.  43. 

But  Emerson  makes  the  product  of  imaginaries  to  be  imaginary ;  and 
for  this  reason,  that  "otherwise  a  real  product  would  be  raised  from 
impossible  factors,  which  is  absurd.  Thus,  y/ —aY.y/ —hz=^ — a&,  and 
^~a  X  —y/  —  6  =  —y^—ah,  &c.  Also,  y/ — a  x  y/ — a  =  —  a,  and 
y/—ax  —  y^— a=+a,  &c." — Emerson's  At,  p.  67. 

From  a  dissertation  "On  the  Arithmetic  of  Impossible  Quantities,"  by 
Mr.  Playfair,  in  the  PhU.  Trans,  for  1778,  p.  318,  we  learn  from  some 
operations  there  performed,  that  he  makes  the  product  oiy/—\  hj  y/—l, 
or  the  square  ofy'— 1,  to  be  —1,  and,  in  another  place,  he  makes  the  pro- 
duct of  y/^1  by  y/U^  to  be y/-l+Z\ 

The  authors  just  quoted  not  only  differ  from  each  other,  but  each  one  seems 
to  be  inconsistent  with  himself.  Thus,  Euber  sa-js,  y/—aXy/—a= — a, 
but  y/ — axy/—b=y/ab,  and  Emerson  says,  y/—axy/—a= — a;  but 
y/—axy/—b=y/—ab.  Now,  the  formula  for  the  product  of  x/ — a 
by  y/—b  ought  to  be  true  whatever  values  may  be  assigned  to  a  and  b. 
Let,  then,  a=b.  Whence,  Euler's  formula  for  the  product  of  y/—a  by 
y/—h^  gives  +y/a»=+a,  and  Emerson's  formula  gives  +^—a^=ay/—l' 
But  they  both  say  that v'— ax v'—a= — «• 

Mr.  Playfair  makes  y/^Xy/l—^=y/—l+z^,  which  we  conceive  is 
not  a  correct  result  when  z  is  more  than  1. 

Let  z=y/2  then  2'=2 ;  whence,  the  above  expression  becomes 
^— 1  X  v/1— 2=x/— 1+2,  or  V— lX\/— 1^=\/1~— Ij  which  result  does 
not  agree  with  his  other  position  unless  he  takes  x/l  =—l,  which  we 
know  would  be  proper  ;  that  is,  when  z  is  more  than  1,  we  have 
yZ-^i X  yi^2^=  — v^— 1+Z3. 


MULTIPLICATION  OF  IMAGINARY  QUANTITIES.         169 

PROBLEM. 

(241.)  Multiply  XTa  by  j^\, 

SOLUTIONv 

Since  |/^=(— 1)2=(— 1)1 =V(— 1)',  we  have, by  the  ordinary 
rule,  V(— l)"xVa=Vl  x  Va=Va.    But  Va=±>i/±Va;  that  is, 

the  4th  root  of  a  is  +4/+V'«,  +|/-Va,  — 4/+Va,  or  — |/ — l^a. 

Now,  to  which  of  these  forms  must  Va,  in  the  present  case,  be  made 
equal  ?     We  have 

ViP^y  X  Va=|A'Plf  X  |/^^=|/i^pr)Va=|/M^ 
Since  v'(— 1)'=— 1.  Therefore,  |/^  x  Va=|/—f^a.  • 

EXAMPLES.* 

1,  Multiply  21^32  by  3V^-^.  ^JW.  —61^6. 


2.  Multiply  4V^— 3  by  94/--I2.  ^      -4w5.  —  216. 

3.  Multiply  —2^"^  by  — 3V^.  ^W5.  — 6V^6. 

4.  Multiply —21/^  by  3 f^^.  -4>w.   +6V6. 

5.  Multiply  l+f/Hiby  1+*^^.  ^tw.  2^^. 

6.  Multiply  1+i^^  by  l-V^^.  Am,  2. 


7.  Multiply  a+i/-6'  by  a+f/-6'.         ^n«.  a»-6'  +  2a6V'-l. 

8.  Multiply  5+ 2|/"^  by  2-V^.  ^ws.  IG-*/"^. 

9.  Multiply  V^  by  V^.  Ans.  2t/^. 


10.  Multiply  2V— 4  by  3V— 16.  Ans.  —12. 

*  A  glance  at  these  examples  shows  that  the  results  are  the  same  as  would 
be  obtained  by  the  following 

RULE. 

To  midtiply  one  quadratic  imaginary  surd  by  another,  multiply  the  quantities 
under  Ihe  radical  signs,  according  to  the  rule  for  signs  given  in  multiplication,  hut 
the  coefficients  of  the  radicals,  axxording  to  the  reverse  of  that  rule ;  thai  is,  marking 
the  product  of  like  signs  minus,  and  unlike  plus. 

This  rule  will  not  hold  when  but  one  of  the  quantities  multiplied  is  an  imagin- 
ary quadratic  surd. 


170  DIVISION  OF  IMAGINARY  QUANTITIES. 


DIVISION    OF    IMAGINARY   QUANTITIES, 

PROBLEM 

f242.)  1.  Divide  |/^  by  |/^. 

SOLUTION. 

_    ,         |/^^   i^axV'^       /a      .         ,     .        . 

We  nave  =-^= :===i/  -:  since  the  unaennary  quantities^ 

i/-b    nxV-1     ^  ^  s       /  1 

cancel. 

PROBLEM 

2.  Divide  -V^  by  —V'^. 


SOLUTION, 


=^=-^Vi. 


-V-h     -VhxV-\  ^ 

EXAMPLE  S- 

1,  Divide  QV^  by  2f'==4.  Ans.  ^VZ, 

2.  Divide  -V~-i  by  -QV^.  Ans,  ^jVl, 

3.  Divide  1  +V~^  by  1— V'^.  Avis.  V^, 

4,  Divide  4  ^-V'^  by  2—V^.  Ans,  1 +V^. 
5*  Divide  1  by  |/— 1,  -4?is.  — V'— 1, 

6.  Divide  a  by  4/af^^.  Ans,  ^^aV—l. 

7,  Divide  2V8— i^^O  by  —S/'^.  Ans,  4/6  +  4*^— 1. 

8,  Divide  b—V^  by  l+i^^.  ^/w.  1— 2f/~2r 

9.  Divide  l4-|/15-(7f'3  +  2V'5)f^^  by  l-^-V^, 


10.  Divide  a  by  hV—l.  Ans,  --?K-1 


^7W.  2—V—9. 


\ 


IKVOLtTTION  OF  IMAGINARY  QUANTITIES.  171 

INVOLUTION    OF    IMAGINARY 
QUANTITIES. 

PB  O  BLEM. 

(243.)  Cube  a-^^\ 

SOLUTION. 

Calling  ^—b^—c,  we  have  (a— c)''=a''— 3a'c  +  3ac'— c'. 

If  now,  we  ascertain  the  value  of  c'  and  c^  we  can  put  these  values 
for  c'  and  c"  in  the  above  expression.  The  square  of  c,  or  of  |/— 6'= 
W^iB  evidently  -b%  and  c'=c\  or  -b'  y-^=^bW—l. 

Hence,  {a—V^y=a'—Sa^bV^—Sab^  +  b^  V^=a^—  3a6'  + 
(b'-Ba'b)i^'^. 

PROBLEM. 

(244.)  To  find  formulas  for  the  powers  of  y— 1. 

SOLUTION. 
Let  the  index  of  the  powers  be  4w,  4w  +  i,  4n+2,  and  4w+8, 
which  comprise  all  positive  integer  numbers. 
Let  a=V—l,  and  we  have 

(|/irr)«»+>=a*M^=a*» .  a'=a'        =-1. 

Thus^  to  obtain  any  power  of  \/—\^  it  is  only  necessary  to  divide 
the  exponent  of  the  power  by  4,  and  the  power  of  j^—\  indicated  by  the 
remainder  will  be  the  result  required. 

EXAMPLES. 

It  Raise  |/—1  to  the  66th  power.         ,  Ans.  —1. 

2i  Raise  4/— 1  to  the  103d  power.  Ans,  —V—l. 

3.  Raise  \/—\  to  the  400th  power.  Ans.  +1. 

4.  Raise  a^— 1  to  the  4wth  power.  Ans.  a*\ 

5.  Raise  aV^— 1  to  the  4w+  1st  power.  Ans.  «*"+'  /— 1. 

6.  Raise  oi/^  to  the  4n  +  2d  power.  Ans.  —  a*"+'. 


172  EVOLUTION  OF  IMAGINARY  QUANTITIES. 

7t  Raise  aV  —  1  to  the  4n+  3rd  power.  Ans.  —  a''^'  /^. 

8.  Square  azhV^.  Atis.  a'— 6±2aV'^. 

9.  Cubeadbf/^.  Ans.  a'— 3a5±[(3a'— 6)V5]f/^. 
10*  Raise  a+V— 1  to  the  6th  power. 


♦  •'  »■ 


EVOLUTION    OF   IMAGINARY    QUANTITIES. 

PEOBLEM. 

(245.)  To  extract  the  square  root  of  a  binomial  surd,  one  of 
whose  terms  is  rational  and  the  other  an  imaginary  quadratic  surd. 

SOLUTION. 

Let  Vx-{-V—y  represent  the  square  root  of  a-\-V—b.  Then 
x-\-2,V—xy — y,  OT  x—y  +  W—xj/^a+V—b;  whence,  by  Theorem. 
(204),  we  have  x-'y=a.    We  know  by  Theorem  (205),  that  if 

^x+V^=ya+V^b,thaiVx---V—y'=  l/a—V'^,     It    is   evi- 

dent  that  the  product  of  |/ic+V— y  bjVx—V^y  which  is  x-\-y,  must 

equal  the  product   of  4/a-\-V—b  by  i/a—V—b,  which  is  Va^-\-b, 

This  gives  x-^y^Va'  +  b.     If  we  extract  the  square  root  of  half  the 
sum  of  x—y  and  a;+y  the  result  must  equal  the  square  root  of  half 

the  sum  of  a  and  f^a'  +  ft;     that  is,  Vx=T .    Also,  if  we 

extract  the  square  root  of  half  the  difference  of  x—y  and  x-^-y^  the 
result  must  equal  the  square  root  of  half  the  difference  of  a   and 

Va^+h;  that  i8,V— y=r .     Whence,  we  have  the  for- 

mula 

Letting  j^a—V—b=:i^x—V—y,  we  would  have 


EVOLUTION  OF  IMAGHNARY  QUANTITIES.  178 

PBOBLEM. 

(246.)  Extract  the  square  root  of  l  +  QV"^, 

SOLUTION. 


Since,  1  +  6V—2  =  1  -{-  V—12,  we  have  by  putting  1  for  a,  and  Y2 
for  b  in  the  general  formula, 


f  2  ^ 

We  could  also  solve  this  by  inspection.  Thus,  since  1  +  6V—2 
=9  +  2  •  SV^—2,  we  see  that  twice  the  product  of  the  square  roots 
of  9  and  —2  is  equal  to  the  second  term,  and  therefore,  3  H-  i^— 2  is 
the  square  root. 

EXAMPLES. 

1,  Extract  the  square  root  of  31 +421^^.       Ans.  1  +  SV^. 

2.  Extract  the  square  root  of  —3+1^—16.        Ans,  1+21^"^. 

3i  Extract  the  square  root  of  4*^^—2.  Ans,  2+V^. 

4i  Extract  the  square  root  of  2  +  4  V — 42.     Ans,  >^14  +  2  V^, 

5*  Extract  the  square  root  of  4/— 1.  Ans,  i4/2+-J-|^— 2. 

6*  Extract  the  square  root  of  —2—2  V—15,    Ans,  \/%—V—5, 

7t  Extract  the  square  root  of  2cdV^l.      Ans.  (1  +^111)4/^ 

8»  Extract  the  square  root  of  8/^.  Ans,  2  +  2t^— 1. 

9.  Extract  the  square  root  of  —V—1.        Ans,  i.f/2— ^V— 2. 

tt'c             etc  ^4cd    — — 
10*  Extract  the  square  root  of  —  —cd -{ ——V—\. 

A71S,  -Vc+V—cd, 
0 

PROBLEM. 
(247  •)  To  extract  the  cube  root  of  a+V^, 

BOMBELLl'S  RULE. 

First  find  Va"  +  6,  then,  hy  trials,  search  out  a  number  c,  awd  a 
square  root  \/d,  such  that  the  sum  of  their  squares  c^-\-dbe  ■=.\yd^  ^  ^j 
and  also,  c^  —  8cd  be=a;  then  shall  e+V—d  be  the  cube  root  vf 
a-\-V—b  sought. 


174  MODULUS. 

PROBLEM. 

(248.)  Extract  the  cube  root  of  2  +  111^^. 

SOLUTION. 


Since,  2  +  11 K— 1=2+ 1^—121,  we  have  a=2  and  6=121; 
whence,  s/a''  +  6=\/125=6  ;  then  taking  c=2,  and  c?=l,  we  obtain 
c*  +  c?=5=\/a'^  +  &,  and  c^— 3cc^=8  — 6=2=a,  as  it  ought;  there- 
fore, 2+1/— 1  is  the  cube  root  of  2  +  llf^— 1.  This  may  also  be 
solved  by  Tartalea's  rule.  Thus  2  can  be  separated  into  two  parts  8 
and  —6,  one  of  which  is  a  perfect  cube,  and  the  other  divisible  by 
3.     Therefore,  r^=8,  or  r=2  and  3s=  — 6,  or  s=  — 2;   whence, 

r+|/-=2  +  y -— ,  or  2+1^—- 1,  the  same  result  as  before. 

EXAMPLES. 

1.  Extract  the  cube  root  of  2  +  2  f^^,  Ans.  —  1  +  f^. 


2.  Extract  the  cube  root  of  2 — V— 121.  Ans.  2 — f  —  1 . 

3.  Extract  the  cube  root  of  SldzSOV^.     Ans.  -3db2|/'^3. 

4.  Extract  the  cube  root  of  — 10  +  9^"^.  Ans.  2  +  V^. 

5.  Extract  the  cube  root  of  —5  — 1^—2.  An^.  1  — |^— 2. 

6.  Extract  the  cube  root  of  —  4  +  lOi/^^.  Ans.  2  +  V^. 

7.  Extract  the  cube  root  of  9  +  25V^^.  Ans.  3  +  V^. 

^  »•  ♦  «.  »i 


MODULUS. 

(249.)  The  modulvs  of  a+/?^^  is  +>fV+^«,  or  the  square 
root  of  the  sum  of  the  squares  of  a  and  ^.  Thus,  +  i/16  +  9=  +  5  is 
the  modulus  of  4  +  3  V—1. 

(250.)  Two  imaginary  expressions  are  conjugate,  when  they  differ 
only  in  the  sign  of  4/— 1 ;  as,  a  +  ^V—l,  and  a—^—l. 
These  conjugate  expressions  have  the  same  modulus. 


MODULUS.  175 

THEOREM. 

(251,)  In  order  that  a-\-^V—l  be  equal  to  zero,  it  is  necessary 
and  sufficient  that  its  modulus  +Va^  +  f^^  be  equal  to  zero. 

DEMONSTRATION. 


If  the  modulus  \/a^  +  i^'  is  not  equal  to  zero,  a  and  ^  will  not  be 
€qual  to  zero  at  the  same  time,  and,  consequently,  a  +  ^V—l  will  not 
be  equal  to  zero.  But,  ifi/a'  +  i?"  is  equal  to  zero,  a  and  ^  must 
each  equal  zero;  whence,  a  +  ^V—l  would  also  be  equal  to  zero. 

THEOREM. 

(252.)  The  modulus  of  the  product  of  two  imaginary  factors  is 
equal  to  the  product  of  their  moduli. 

DEMONSTRATION. 

The  product  of  a  +  ^V^l  and  a'-\-^'V-^l  is  {aa'-§§')^ 
^(i^'  ^^a')V—\,  and  the  modulus  of  their  product  is,  therefore,  equal 
to   ^{aa^  -  §^'y  +  {a§'  +  §afY  =  Va'a''  +  ^'§^'  +  a'^^'  +  ^'w'  = 

Va'''  +  §'\     Q.  K  D. 

THEOREM. 

(2 53.)  The  modulus  of  the  quotient  resulting  from  the  division 
of  one  imaginary  quantity  by  another  is  equal  to  the  quotient  of  their 
muduli. 

DEMONSTRATION. 

Let  a" -{-^''V—l  represent  the  quotient  of  a-\-§V—l  by 
a'  +  |9Y— 1,  since  it  can  be  easily  proved  that  the  quotient  must  be 
of  this  form.     Hence,  we  have, 

a  + 19|/^=  (a' +  (^ Y-T)  (a"  +  ($' Y~-I1). 
By  the  last  Theorem,  we  have  |/^M^'=  Va'^-[-tifWa^''+§"\ 
We  may  consider  ^/a"'*  +  ^""^  as  the  quotient  arising  from  dividing 

VcT^  by  |/a'^  +  <?'» ;   that  is,  J^ir^^'z=:  ^""'^^^  .     Q.  E,  i>. 


176  EXAMPLES   IN   RADICALS. 


THEOREM. 


(254.)  In  (wder  that  the  product  of  imaginary  factors  be  equal 
to  zero,  it  is  necessary  and  sufficient  that  one  of  the  factors  be  equal 
to  zero. 

DEMONSTRATION. 

It  can  be  easily  shown  that  the  product  of  two  or  more  imaginary 
fectors  must  be  of  the  form  a-\-?V—l. 

In  order  that  this  product  may  be  equal  to  zero,  we  have  seen  that 
its  modulus  must  be  equal  to  zero ;  but  this  modulus  is  the  product 
of  the  moduli  of  the  several  factors,  and  these  moduli  are  real  quanti- 
ties, and  consequently  their  product  can  not  be  zero,  unless  at  least 
one  of  these  moduli  be  also  equal  to  zero ;  but  when  a  modulus  is 
equal  to  zero,  its  corresponding  imaginary  factor  must  also  be  equal 
to  zero.     Hence,  the  Theorem  is  true. 

MISCELLANEOUS    EXAMPLES    IN    RADICALS. 

1.  Extract  the  cube  root  of  20  +  14^2.  Ans.  2+f2. 

2.  Simplify  L^^^Q-^.  Am,  -{5  +  VlO). 

22-74/10 

3t  Extract  the  square  root  of  —1 +4V'^.         Ans,  2+1^—5. 
4.  Simplify  3 via" +  2l/2a.  Ans,  6V2a. 


5,  Simplify  VSa'b  +16a*—Vh*  +  2ab\       Am.  (2a — 6) l/2a  +  b. 


6.  Simphfya./^3  +  276^-  Am, -Va  +  2K 


7.  Simplify  4/4a'6'—20a'6'  +  25a6*.  Am,  {2d'—b)bVa, 

8.  Simplify  V81-2V24+f2"8'+ 2^/63.  Am,  8f7-V3. 

9.  Sunplify  4/12+2m  +  34/76  +  9V48.  Am,  b^Vl. 


10.  Simplify  .  Ans,  ——Vx, 

--     ^.      ,.^    ct—b  Vac'  .        Vac 

11,  Simphfy r  •  — .  Am,  ——j-. 


EXAMPLES  IN   RADICALS.  177 

a-\-b     /a—b  J        .  A+^ 

12.  Simplify  _±_/__.  ^-Y^- 

13.  Simplify  V2  X  Vi  X  V3.  ^«s.  "x/'^i. 

14.  Simplify  1/4  X  "v/3  X  V6.  ^rw.  'V3981312. 

15.  Find  the  sum  of  Vi  and  VS-  ^^^-  F^^. 

16.  Find  the  difference  of  V|  and  VV--  ^^^-  f  j V75. 

17.  Find  a  factor  that  will  make  a"'  rational.  ^%5.  a\ 

18.  Simplify -T= r:.  Am.  dVlsVd. 

V5  4-^3 


19.  Extract  the  square  root  of  b€-^2bVbc—b^ 


Am.  db(6+V6c-6'). 


20.  Extract  the  square  root  of  wp+  2^"— 2mVwi?+wi". 


Am.  ±(t/w23  +  m'— m). 


21.  Simplify  ^16  +  30|/-l+|/l6-30V/~l. 

Am.  ±6l/^. 


22.  Reduce  i/l6  +  30V— 1  +i/l6  — 30i^— 1  to  its  simplest  form. 

^TIS.   ±10. 

23.  Simplify  |/ 3 1  + 1 21/'^  + 1/— 1  -h 4 V-^. 


24.  Reduce  i/s  1  +  12V'— 54-|/— 1+4V— Stoits  simplest  form. 

^W5.  rb4. 


25.  SimpUfy  ^bc-{-2bVbc-b'  -{.^bc-2bVbc-b' 


Am.  ±:2Vhc—b\ 


26.  Reduce  A/bc  +  2bVhc—b'  ■\-A/bc—2bVbc-b''   to   its   simplest 
form.  Ans.  ±2&. 

28.  Divide  2i^3  x  \/l  by  iV2  x  V3.  -4w«.  4 V288. 

12 


178  EXAMPLES   IN   RADICALS. 

29.  Divide  ^f/j  by  |/2  +  3  V^.  Am.  J^. 

30.-  Multiply  |/2  X  VS  by  Vi  x  Vj-  ^^^'  '^^8. 

31.  Multiply  V|  X  Vi  by  "v/6~  ^n*.  *V^. 


32.  Dividei/Vi  X  2 V3  by  i/4V2  xf  3-  ^7i5.  ^Vf. 

33.  Divide  1  by  Va-\-Vb. 

Va'-V^b  +  Vab'-Vf' 

Ans.  ; . 

a — 0 

31.  Multiply  41/f +5^1  by  i/i  +  2Vl.  Ans.  ^+^-^^2. 

35.  Divide  Va  +  V6  by  Va-V6. 

a-{-b  +  2Vab  +  2Varb  +  2Va^ 


Ans. 


a—b 


36.  Cube Ans.  1. 

2 

37.  Simplify^^?^^-^^.  Ans.  4-V3, 

38.  Simplify   )  2(2FV3  f  ^  ^^^^  ^rlr-Vi. 

(2V2(3)^f  3«* 

39.  Simplify  ./  )  (i^jtl^  (  '.  ^,,5.  i/i(\VQ'hV2i). 

^    (   2i/2-(i)2    ^  ^ 

—  3 1^2  4-2  — 

40.  Find  the  sum  of  6^/2  — 1  and =:.  Ans.  3  +  8t/2. 

_5^-3t^2 

'V'f4+Vl2        V6—V4c 
41*  Find  the  difference  between  — = =-  and  -^^ 1_. 

VI-  V6  Vs  +  V2 

Ans.  SV2  +  4V2i  +  4VS. 

42.  Divide  2  V'S  x  4Vi  by  4Vi  x  4V4.  Ans.  1. 

43.  Extract  the  square  root  of  1  —  VlS.  Ans.  i|/26-^f2. 

44.  Extract  the  square  root  of  H-fVi  — ^V^- 

Ans.  iV^3-iV6. 


CHAPTER  X. 
EaUATIONS. 

(255.)  An  equation  is  an  algebraic  statement  denoting  the 
equality,  in  value,  of  two  algebraic  expressions;  as,  ax-\-h=c, 
6a;'*  +  2a;='7,  and  aa?— 6=0. 

(256.)  The  absolute  term  of  an  equation,  is  that  term  which  is 
completely  known,  or  is  considered  as  known,  and  which  is  equal  to 
the  sum  of  all  the  unknown  terms.  Thus,  in  the  equations  6a: =9, 
4a;'^  +  3a;— 7=0,  and  ax=b—c',  9,  7,  and  (6— c)  are  the  absolute 
terms, 

(257.)  ThQ  first,  or  left  hand  member  of  an  equation,  is  the  part 
of  the  equation  which  precedes  the  sign  of  equality. 

(258.)  The  second  or  right  hand  member  of  an  equation  is  the 
part  of  the  equation  which  follows  the  sign  of  equality. 

THEOREM. 
(259.)  Any  term  of  an  equation  may  be  transposed  from  one 
member  to  the  other  by  changing  its  sign. 

DEMONSTRATION. 

Let  aa;  +  6=c  be  an  equation.  This  equation  may  assume  the  fol- 
lowing form  ax-\-b=^c-\-b~b.  It  is  evident  that  we  should  still 
have  a  correct  equation  if  we  should  omit  +6  in  both  members. 
Thus  ax=zc—b. 

Again,  let  ax—b=:d.  Because  ax—b=d+b—b,  we  have 
ax'=d-^b.  These  results  show  that  if  we  transpose  any  term  of  an 
equation  from  one  member  to  the  other,  its  sign  must  be  changed  in 
order  not  to  destroy  the  equality. 

The  truth  of  this  Theorem  may  be  proved  as  follows  : 

Letting  aa;  + 6 =c  and  subtracting  -\-b  from  both  members,  we 
have  ax-\-b=zc 

+b^b 
ax  '  =c— 6, 


180  SIMPLE   EQUATIONS. 

or  by  adding— 6  to  both  members,  we  have 

ax-^b=c 
-b=-b 

ax       z=.c—h. 
Letting  aa;— 6= (^  and  subtracting    —6  from  both  members,  we 
have  ax—b=d 

-b=-b 
ax       =zd  +  bj 
or  by  adding  +  ^  to  both  members,  we  have 
ax—b=d 
-\-b=+b 

PROBLEM. 

(260.)  Transpose  all  the  known  terms  of  a?*4-6a;"-~4ar— 4+rf 
—6=0  to  the  second  member. 

S  OLUTION. 

Transposing —4,  +c?,  —bhj  changing  their  signs,  or  by  adding 
+  4— c?  +  6  to  both  members,  or  by  subtracting  —  4+<?— 6  from 
both  members,  we  have  x^-\-Qx^—4:X—^—d  +  b. 

EXAMPLES. 

1.  Transpose  in  6a; +  4= 4a; +  8  the  terms  4-4,  and  4a?. 

Ans.  2a; =4. 

2.  Transpose  in  Sa;" — 2a;  +  3=4a;''— 4a;+l  the  terms  +3,  4a;',  and 
—4a;.  Ans.  x^  +  2x=  —  2. 

3*  Transpose  in =-H —  the  terms and-.     Ans,  -=-. 

X     y     x     y  y         X  X     y 

4    m_  .4334,  3 

4.  Transpose   m  — -— -|-— — -=_ — _j the  terms   and 

x+a     x  +  y     x  +  a     x-\-y  x+y 

Ans. 


^  +  «  *  x-\-a     x+y' 

f,    _,  .     4a;+4      a—b     ^     2a;+l     a-\-b 

5.  Transpose  m -— — ^  +  6=— — ^  +  — ^  +  5,   the    term 

bx-^b     x--d  5a;  +  6     x—d       * 

a—b  2a;  +  l  2ar  +  3        2a 

—  — :t,  +6  and -.  Ans.  — ^-=: -— 1. 

x—d  hx-^b  hx^-b     x—d 

6.  Transpose  in  Ya;*— 6a;'4-6=-~a;*-f-8  the  terms  -}-6,  and  —a;'. 

Ans.  1x*—5x'=2, 


SIMPLE  EQUATIONS.  181 

7,  Transpose  in  4  Vx+  Vy=Vx-\-dVy  the  terms  |/y  and  ^x. 

Ans.  sVxz=2Vy. 

2i/3  4-3i^2  24/3"— 31^2 

8.  Transpose  in ——^^^ +  4= +  9  the  terms   +4, 

^  2x  2x 

2x  X 

9t  Transpose  in  aVx-\-cVd=ihVx-\-hVd  the  term  cVd  and  hVx, 

Ans.  {a-h)Vxz=i{J)-c)Vd. 

*A    m^  .    9a;  +  20     4a;— 12     x   .     ^        x 

10.  Transpose  m  — — — =— j-+t  the  term  -. 

5     4a;— 12 

Ans.  -= — . 

9      6a;— 4 

11.  Transpose  m  — — rT=«  the  terms — -  and  -. 

^  21         4a;— 11     3  4a;— 11         8 

16       a;  +  8 
Ans. 


21     4a;-ll 


^«    ^  .    20a;     36     6a;  +  20     4a;     86^,     ^  36       ,  4a; 

12.  Transpose  m  -77^+777+7^ t^=-t-+777  *^®  *®"^s  777  ^^<i  "T"* 

^  26      26     9a;— 16      6      26  26  6 

5a;  +  20     ^ 


9a;— 16 
THEOREM. 
(261 1)  Any  equation  containing  fractions  may  he  cleared  of  these 
fractions  by  multiplying  both  members  by  the  least  common  multiple 
of  the  denominators  of  the  fractions. 

DEMONSTRATION. 

ax       ex     d      c 

Let  -r-  4- -^,=  -5-1 1-6  he  an  equation  containing  fractions. 

b       ab^     a^     a  ^ 

The  least  common  multiple  of  the  denominators  is  a'6'. 

Multiplying  both  members  of  the  equation  then  by  a'ft",  we  have 
a^bx'  +  acx=b'd -{- ab^c+a'b^,  an  equation  which  contains  no  fractions. 

It  is  evident  that  the  same  method  of  proceeding  would  produce  an 
equation  containing  no  fractions,  whatever  might  be  the  number  of 
fractions  in  the  equation  to  be  cleared. 

Corollary. — To  clear  an  equation  of  fractions,  we  may  multiply 
by  any  common  multiple  of  the  denominators. 

PROBLEM 

(262.)  1.  Cleai  ^a;*4-ia;=8  effractions. 


182  SIMPLE    EQUATIONS. 

SOLUTION. 

The  least  common  multiple  of  the  denominators  is  12. 
Multiplying  both  members  by  12,  we  have  3a;''-f-2a;=96. 
We  might  also,  multiply  by  24,  or  any  multiple  of  12. 

PROBLEM 

2.  Clear  a; +  -+-—+ -T-H — -  =  146  of  fractions. 
2      4       7      14 

SOLUTION. 

Multiplying  both  members  by  56,  the  least  common  multiple  of  the 
denominators,  we  have  66a;  +  28a;-|-42a;  +  16a;-f4a;=146  •  66. 

PROBLEM 

^    ^.       3x     x  —  l     ^       20a;  +  13     ..       . 
S.Clear— -— =6a; : of  fractions. 


SOLUTION. 

Multiplying  by  4,  we  obtain  3a;— 2a;  +  2  =  24a;— 20a;— 13. 

It  must  be  carefully  noted  that  in  the  numerators  a;— 1,  and 
20a;-f-13,  the  x  and  20a;  are  both  positive,  the  negative  sign  belonging 
to  the  whole  fraction. 

Corollary. — Hence,  we  see  that  when  a  denominator  is  removed 
from  a  negative  fraction  by  multiplication  or  division,  every  term  of 
the  resulting  numerator  must  have  its  sign  changed. 

PROBLEM 

2i/— a;  59— 2a; 

4.  Clear  a; — =20 — of  fractions. 

23— a;  2 

SOLUTION. 

The  least  common  multiple  of  the  denominators  is  2(23— a;)  = 

46— 2a;.     But,  in  order  to  save  labor,  we  shall  first  multiply  by  2, 

and  then  by  23— a;. 

4y — 2a; 
Multiplying  by  2,  we  get  2a;— ^^ =40  -59  4-2ar. 

Zo  — X 

Let  US  simplify  this  equation  before  multiplying  by  23— a;.  This 
can  be  done  by  dropping  2a;  from  both  members,  and  adding  40  and 

—69  together,  whence  results =  —  19. 

^         '  23— a; 

Now  multiplying  both  members  by  23— a;,  we  obtain  — 4y4-2a;= 

—  19  •  23-hl9.r,  or  2a;— 4y  =  19a:  — 437. 


SIMPLE  EQUATIONS.  188 


EXAMPLBS. 

1.  Clear -=5  +  2  of  fractions.  Ans,  x=y+10, 

5     5 

2.  Clear  6a;— ^=-?^— J^+37  of  fractions. 

3       4        6 

Ans.  60a;— 4y=9a;— 16y+444. 

3.  Clear4—?^^=y—l'7f  effractions. 


f- 


t 


Ans,  24— y+a;=6y— 106. 

.    ^,       Ta;— 21  x     ^     3a;— 19    ^^ 

4,  Clear  — 1-  y — -  =  4  4 —  of  fractions. 

o  o  2 

Ans.  Ya;— 21+6y— 2a;=24+9a;— 5Y. 

5.  Clear  ?^-?^=?l^-^^  effractions. 

^  o  4  lo 

Ans.  16a;  +  8y— 18a; +14= 12y  + 36— 4a;— 6y. 
6«  Clear  J(a;+y)  + 6 =y  of  fractions.  Ans.  a;  +  y+18=3y. 

»    ^,  3a;— 3     ^       3a;— 6    ^^ 

7.  Clear  5a; ;r-= 2a;  H —  of  fractions. 

a;— 3  2 

Ans.  10a;'— 30a;— 6a;  +  6=4a;'— 12a;  +  3a;'— 16a;+18. 

8.  Clear --\ =— -  of  fractions. 

a;+l        X         6 


Ans.  6a;'  +  6a;'  +  12a;  +  6  =  13a;»+18a;. 

21^3; +  2     4 )/x  — 

9.  Clear == :=- of  fractions.     Ans.  2a;  +  2|/a;=16— ar. 

4+l^a;         Vx 

10.  Clear  i/?i^+2i/-^=6i/-^  effractions. 

r       X  '  x-\-a       '    x-\-a 

Ans.  x-{-a  +  2Vax=h'x, 


SIMPLE  EQUATIONS. 

(263.)  A  Simple  Equation  is  one  in  which  the  exponents  of  the 
unknown  terms  are  ±1,  or  =t,  a  proper  fraction  whose  numerator  is 

one  ;  as,  6r  +  a=6,  1x^-\-y^  —  S,  ar-'  +  c=rf,  4a;~« +5  =  9,  <fec. 


184  SIMPLE  EQUATIONS. 

SIMPLE    EQUATIONS   CONTAINING   ONE    UNKNOWN 
QUANTITY. 

PROBLEM. 
(264.)  To  solve  a  simple  equation  containing  but  one  unknown 
quantity,  that  is,  to  reduce  it  to  its  simplest  form. 

RULE. 

1.  Collect  the  terms  as  much  as  possible  by  addition  and  transpo- 
ntion, 

2.  Clear  the  equation  of  fra/^tions. 

3.  Transpose  all  the  unknown  terms  to  the  first  member,  and  the 
known  to  the  second  member, 

4.  Collect  all  the  unknown  term^  when  not  already  done^  into  one 
term,  and  put  the  second  member  in  its  simplest  form. 

5.  Divide  both  members  of  the  equation  by  the  coefficient  of  the  un^ 
known  Quantity,  and  the  resulting  equation  will  be  the  simplest  form 
of  the  given  equation,  and  will  indicate  the  value  of  the  unknown 
quantity. 

DEMONSTRATION. 
It  is  evident  that  the  operations  indicated  in  the  rule  are  such  as 
will  not  destroy  the  truth  of  the  equation,  nor  will  they  change  the 
value  of  the  unknown  quantity.  Hence,  the  value  of  the  unknown 
quantity  in  the  final  equation  will  always  be  the  same  as  in  the  equa- 
tion from  which  it  results.    . 

PROBLEM 

(265.)  1.  Solve  3a:-4=7a?-16.  (1). 

SOLUTION. 

— 4a;=  —-12  {2)=Eq.  ( 1)  transposed  and  added. 

x-Z.  (3)=(2)-^_4. 

PROBLEM 

2.  Solve  ax^c=d—bx.  (I). 

SOLUTION. 

ax-\-bx=:c-\-d  (2)  =  (1)  transposed. 

{a-\-b)x=c-\-d  (3)=(2)  added. 


SIMPLE   EQUATIONS. 


186 


PROBLEM 

8.  Solve  ax^-\-bx=dx'-{-cx,  (1). 


SOLUTION, 


ax-^-  b=9x-^c 

ax—9x=  c—h 

(a— 9)a;=  c—h 

c-h 


(2)=(1)-^. 

(3)  =  (2)  transposed. 

(4) =(3)  added. 


—  (5)=(4)-^(a-9). 


PROBLEM 

,    ^  ,       20a;      36     6a;  +  20     4a;     86 
4.  Solve  —  +_+^_^=-+_. 


SOLUTION 


This  equation  may  be  put  in  the  following  form, 

Dropping  equals  from  both 


4a;     36     5a;  + 20     4a;     ^     36     ,  . 

y+25-^9S3r6=y+^+25-  (^) 

members,  we  get Tr~ 

6a;+ 20=18a;— 32 
— 13a;=— 62 
a;=:3 


(2). 

(3) =(2)  cleared  of  fractions. 
(4) =(3)  transposed. 
(6)=(4)--13. 


PBOBLEH 

\,  ■.     ..     .  2ii;     5x    X     X      „,  _ 
6.  Solve  nx+—-^—+-+—=Z\5. 


(1). 


SOLUTION. 

264a;+16a;  +  30a;+4a;+a;=316  x  24  (2)=(1)  X  24. 

316a;=316  x  24  (3)  =  (2)  added. 

a;=24  .  (4)=(3)~315. 

Eemabe. — ^No  multiplication  should  be  performed  -unless  there  is  a  necessity 
for  it.  If  we  had  multiplied  315  by  24,  we  would  in  this  case  have  performed 
work  which  was  unnecessary. 

PROBLEM 


6.  Solve  V4a;  +  16=12.  (1). 


186  SIMPLE  EQUATIONS. 

SOLUTIO  N. 

We  must  first  free  this  equation  of  the  radical,  which  can  be  done 
by  squaring  both  members.     We  learned  in  Involution  that  to  square 
the  square  root  of  a  quantity,  we  have  only  to  omit  the  radical. 
4a:+ 16  =  144  (2)  =  (1)'. 

4a; =128  (3)  =  (2)  transposed. 

X-  32  (4)=(3)-T-4. 

PROS  LEM 


7. 

Solve  |/12  H-ir=2  4-  f  a:.             (1). 

SOLUTION 

. 

12  4-a:=:4H-44/^-l-a; 

(2)-(l)'. 

8=4|/5 

(3)— (2)  transposed. 

1=.Vx 

(4)=(3)-4. 

4:=X 

(5)=(4)'. 

PROBLEM 

8. 

Given  ^x—a=Vx—lVa  (1)  to  find  the  value  of  a?. 

SOLUTION 

. 

x—a=x—Vax+ , 
4 

(2)=(1)\ 

.—    5a 
i/ax=~ 

25a'' 

ax= 

16 

- 

25a 

PROBLEM 

9. 

Solve  ^^  +  2'     ^^~+^^. 

^x+4     Vx+e 

(!)• 

SOLUTION 

. 

a;  +  34far  +  168=ar  +  42|/a;+162  (2)=(1)  x  (Va;  +  4)(Va?+6). 

16  =  8f/i"  (3)=(2)  transposed. 

2=zVx~  (4)=(3)-f-8. 

4=x  (5)  =  (4r. 

PROBLEM 

10.  Solve  V2^T3  + 4=7.(1). 


SIMPLE   EQUATIONS.  187 


SO  LUTIO 


V2a;  +  3=3  (2)  =  (1)  transposed. 

2a;  +  3=2Y  (3)=(2)'. 

2a;=24. 
a;=12. 


PROBLEM 


11.  Given  -+^=i/— +^^  .  A(l)  to  find  the  value  of  x. 


a'x       X 

SOLUTION 


a^'^ax'^a'-a'^V  a V  +  a;* 

(2)=(1)'. 

12/4          9 
x''^ax~y    aV  '^  X* 

(3) =(2)  transposed. 

12/49 

(4)=(8)xir. 

14       4       4       9 
x''^ ax'^  a'~  a''^  x* 

(6)=(4)'. 

4_8 
ax~  x^ 

(6) =(5)  transposed. 

1_2 

a~x 

(7)=(6)xf- 

x=2a 

(8)=(7)x<w. 

PROBLEM 

a/cl  ~~~  \/ Oi  ~~-  X 

12.  Given  -^ =a  (1)  to  find  the  value  of  x. 

^a-{-Va—x 

80  LUTIO  N. 

Kendering  the  denominator  of  the  fraction  which  forms  the  first 
member  rational  by  multiplying  both  numerator  and  denominator  by 
|/a—  \^a—x,  the  equation  becomes 


^a—^Va^—ax—x  .  . 
=:a             (2). 


—{a-\-l)x-^2a=i/4a''—4ax         (3)  =  (2)  xa?  and  trans. 
(a  +  l)V— (4a''  +  4a)a;  +  4a'^=  4.a''—4:ax         (4)=:(3)'' 

(a  +  l)''a;-4a''-4a=— 4a  (5)  =  (4)~a; 

(a  +  lYx=4a'' 

_     4a^     _/  2a  V 


188  SIMPLE  EQUATIONS. 

PROBLEM 


13.  Solve  Vx  +  2  +Vx=-=,  (1) 

N  Vx  +  2 


SOLUTIOl? 


x+2-hVx'  +  2x=4:  (2)=(1)  xVx+2 

Vx''-\-2x=2—x 
aj*  +  2a;=4— 4a;+a?* 

6X  =  4: 

FBOBLEM 

,>    ai      ^     8a;  +  25     17— 6a;     ^,    .        9a;+40      ,,. 

14.  Solves ;p- __=2J^+a; 1—.    (1) 

SOLUTION. 

20-8.-26-«?=?^=8J  +  4.-?^  (2)=(1)X4 

68— 24a;     ,^,      .        9a;+40         ,^.      ,^, 
5 =131^4- 7a; ^^ —        (3)=(2)  transposed. 

-68  +  24a;=118  +  63a;--ii^^tl5£         (4)^.g^x9 

2 

81a;+360 

=39a;+186  (5j=(4)  transposed. 

81a;+360=78a;  +  372 

3a;=12 

«=4. 

PROBLEM 

15.  Given  — - — :  — - — : :  7  : 4  to  jfind  the  value  of  x, 

2  4 

SOLUTION. 

In  a  proportion,  we  have  the  first  term  divided  by  the  second,  equal 
to  the  third  term  divided  by  the  fourth. 
By  performing  these  divisions,  we  have 
10a;+8_7 
18-a;"~4 
40a;+32=126--7aj 
47a;=94 
«=2 


SmPLE  EQUATIONS.  189 

PROBLEM 

4bX  4-3 

16.  Given —  :  1 : :  2.r  + 19  :  3a;— 19  to  find  the  value  of  a?.    ^ 

6a;— 43 

SOLUTION. 

Putting  this  proportion  in  the  form  of  an  equation,  we  have 
4a;  H- 3      2a; +  19 

6^=43=37=19-      ^^'^"^^     ^^    ^^*^^^» 
we  have      12a;''4-9a;— 76a;-67=12a;''+114a;—86a;— 817 
— 95a;=  — 760 
a;=8 

PROBLEM 

17.  Given  (a+a;)(6+a;)— a(6+c)=—  +a;'(l)  to  find  the  value  of  a?. 

S  OLUTION. 

ah'\-(a+b)x+a^'-cLb^ac=—+x^  (2)=(1)  expanded. 

(a+h)x=—+ac  (3) =(2)  transposed. 

/    ,  rx  _«'c  +  a5tf  ,._..)  with  tenns  of  2d 

^        ^  ~       b  ^  ''"^  M    member  added. 


ac 


(a + b)x=z~(a  +  b)         (5) = (4)  factored. 


EXAMPLES. 

1,  Given  8a;+7=52  — 7a;  to  find  the  value  of  a;.  An^,  x=S. 
2*  Given  18a;— 13= 6a; -f  36  to  find  the  value  of  a;.  Ans.  a; =4. 
3*  Given  19a;+13=:59— 4a;  to  find  the  value  of  a;.     Ans.  a;=2. 

X 

4*  Given  3a;  +  4— -=46— 2a;  to  find  the  value  of  a;. 
3 

Ans,  a;=9. 
5,  Given  x^  +  15a;=35a;— Sar'  to  find  the  value  of  a;.  Ans.  x=5. 
6«  Given  4a;  +  36=5a;-f-34  to  find  the  value  of  a;.  Ans.  a; =2. 
7.  Given  3a;''— 10a; = 8a; +  3;"  to  find  the  value  of  x.  Ans.  x=9. 
9»  Given  Sax—4cab=2ax—6ac  to  find  the  value  of  x. 

Ans.  a; =46— 6c. 


190  SIMPLE  EQUATIONS. 

9.  Given  ax^  +  abx=cdx  to  find  the  value  of  x. 


.  cd—ab 

Am,  x=: , 

a 


,  X     X 

10*  Given  rr— 7  =--f- to  find  the  value  of  a:.  Ans.  a?=15. 

6     3 

XXX 

1 1 »  Given  -4--=-+7to  find  the  value  of  x.  Ans,  a;=  1 2. 

Z        O        He 

/g K  284 X 

12t  Given \-6x= to  find  the  value  of  x, 

4  5 

Ans.  x=:9 
J  J ^      jg 2; 

13.  Given  x-] — = — - —  to  find  the  value  of  a:.    Ans,  x=5, 

o  2 

II*  Given  3a?  H — =5H to  find  the  value  of  x, 

5  2 

Ans.  a;=7. 

«.r    r,'        ^,      8a;— 11     6a?— 5     9l—1x^    n    .  ^t.       i        i? 

15,  Given  21  -\ — — = — 1 —  to  find  the  value  of  x. 

16  8  2 

Ans.  a? =9. 


g/p 4  2g ^/p 

16t  Given 2= — \-x  to  find  the  value  of  a?. 


Ans.  x—4, 

17.  Given  ~_-  +  10=-— -  +  11  to  find  the  value  of  a?, 
o      4  3      2 

Ans.  arrr  12. 

/g  I   1  2a; 3 

18i  Given  — - — 1-3= — - —  to  find  the  value  of  a?.      Ans.  x=9, 
5  3 

'7a;4-2  5x 6 

19t  Given  — h  5a?= 28  H z —  to  find  the  value  of  a?. 


Ans,  a;=4. 

Sx4-4:  22 — X 

20.  Given  — -^  +  2a;  =— - —  + 16  to  find  the  value  of  a?. 

6  5 

Ans.  x=1. 

9t     n-        7a;— 8     16a; +  8     ^       31— a;  ^    c  j  ^v        i        ^ 

21.  Given  -— - — I — — =3a; —  to  find  the  value  of  a;. 

11  13  2 


Ans.  x=9. 


22.  Given  4a; — =.16 to  find  the  value  of  a;. 


Ans.  x=S. 


SIMPLE   EQUATIONS.  191 

^^    ^.  21— 9x     bx  +  2     61     2a;  +  5     29  +  4a:  ^     «    ,  ^,       -+• 

23.  Given  «  +  — Q~~='\2 3 \2 

value  of  a;.  Ans.  x=:5. 

„.     ^.        31  +  4a;     3a;  +  47     3a:— 19        .,     16  — 10a?      6a;+20 

24.  Given -^ ^=41^  +  -^^ ^ 

to  find  the  value  of  a;.  Ans,  xz=il1. 

25 1  Given 5 =- to  find  the  value  of  a;.     Ans.  x— — - — . 

XX  4 

26.  Given    °       —      W    n      /=i4^^^^  ^ 
of  a;.                                                                                ^^.  a;=2|. 

««    ^.        4a;-l7     3|-22a;  6  j  ,      a;'  i         xj   ,   ^v         , 

27.  Given — ^-—- — =a; — -{1— rr}-  to  find   the   value 

9  83  X  \         54  ) 

of  a;.  Ans,  a;=3. 

28.  Given  \  \  ?a?4-4  [  --^1?=^  -j  ?-l  [  to  find  the  value  of  a;. 

2*3  •  3  u.\X         ) 

Am,  a;=3. 

29.  Given  3*25a;— 5*007— a;=0-2—0-34a;  to  find  the  value  of  ar.      * 

Ans,  a;= 2-0104247. 

7*53aj  2aJ  x 

30.  Given  -—- 100=—  —  3-86—  -  to  find  the  value  of  x, 

18  5  6 

Ans,  a;=4-519-675.    -~ 

31.  Given  — n  +  /    .  rxs  +     /    .  rxa  =3ca;+—  to  the  value  of  a;. 

.  ah 

Ans,  x 


32.  Given r-~4 to  ^^  t^e  value  of  a;. 

a+bx     d  +  ex 


Ans,  x- 


af—cd 
ce—hf' 


t«  Given  Va'  +  c =4/^77 r  to  find  the  value  of  a;, 

y  d(x-\-g) 


d{x-\-g) 

Ans,  a;= —g. 

dX/a^+c 


34.  Given  Va+a;='Va:'  +  6aa:+&'  to  find  the  value  of  a;. 

Ans,  x=—- — . 
3a 


192  SIMPLE  EQUATIONS- 

bx        (Sbc-}-ad)x        Sab   _(8bc—ad)x  ^ 5a(2b—a) 
^*      ^^^^  2b^^'~~2ab(a-{-b)      d^^"  2ab{a-b)       ~~a^-b^ 

to  find  the  value  of  a;.  Ans.  x=-—^ =-^, 

36.  Given -= :; —z — :; —  to  find  the  value  of  a?. 

a?— 1         x-\-l  x^—1 

Ans,  «=— IJ. 

X. 

Ans.  x=8. 


9a; +20     4a;— 12      x 

Zl*  Given  — -- — =-- +-  to  find  the  value  of  x. 

36  6a;— 4      4 


38*  Given =-  to  find  the  value  of  a;. 

21  4a;— 11     3 

Ans,  a;=8. 

39.  Given  — - — +  — —  =  — — -  to  find  the  value  of  x. 

9  Da;+3  3 

Ans,  a; =4. 

H^40.  Given  — 14  =  ^^~^  — 15  to  find  the  value  of  x, 

*  a;  +  2  2a;— 2 

Ans.  x=2. 

/p 2     2Vx 

41*  Given  — =-=— -  to  find  the  value  of  a;.  Ans.  x=6» 

|/a;         3 


42.  Given  a;-H/2aa;+a;''=a  to  find  the  value  of  a;.        Ans.  x—-. 

4 


43.  Given  2Va*+a;'=4(a— ^a;)  to  find  the  value  of  a;. 


J  3a 

Ans.  x=z—. 


44*  Given  a-\-x=A/a^-\-xVb^+x''  to  find  the  value  of  x. 

b'^4a' 


Ans.  x=- 


4a 


45.  Given  y'3a;— 1=2  to  find  the  value  of  a;.  Ans.  a;=f. 


46«  Given  |/a;4-a;'=a;  +  i  to  find  the  value  of  a;.  Ans.  x=:^. 


47.  Given  \/3a;  + 13— 4=0  to  find  the  value  of  x.      Ans.  a?==l7. 


48.  Given +  a;=g  +  2a;  to  find  the  value  of  x. 

Va—x 

Ans.  a;=l— a. 


49.  Given  y4-\-Vx*---x^=  x—2  to  find  the  value  of  a;. 

Ans.  a;=2|. 


SIMPLE  EQUATIONS.  '     198 


50.  Given  (2+a;)2+a:2=4(2+a;)"2  to  find  the  value  of  a;. 

Ans. 

51.  Given  3/2^6  +  3  =  15  to  find  the  value  of  x,     Ans.  x=5. 


Ans.  x=^. 


52.  Given  |/a;  +  3=f^21  +a;  to  find  the  value  of  rr.        Ans,  a; =4. 


a 


53.  Given  x-\-Va—x—  to  find  the  value  of  a?. 

Va—x 

Ans.  a;=a— 1. 

54«  Given -^^ -/=— —  to  find  the  value  of  a;. 

3a;         3a; +  2       11a; 

Ans.  a;=yYT' 

CL  C 

55*  Given  a;+T«-|-r-a;=w»  to  find  the  value  of  a;. 

O         0 

^     '  bm 

Ans.  X 


56f  Given  \/a-^x-{-Va~-x=Vax  to  find  the  value  of  a;. 

Ans.  x=- 


a" +  4* 


57*  Given  i/— -— +  |/— f_  —a/      ^     to  find  the  value  of  a;. 
"   a4-a;      ~   a—x     ^   a^—x^ 


Ans. 


58*  Given  \fx-\-az=.c—Vx-\-h  to  find  the  value  of  a;. 

Ans.  a;=l 1 

V       2c      / 

59*  Given  i^a+Vx^zVax  to  find  the  va%ie  of  x. 


Ans.  a;=: — =^- 


{Va-\f 


60*  Given  |/a;— 16=8— f^a;  to  find  the  value  of  a;.      Ans.  a;=25. 


6)t  Given  |/a;-f  40=10— Va;  to  find  the  value  of  a;.      Ans.  a; =9. 


62.  Given  4/a;— 24=V^a;— 2  to  find  the  value  of  a;.      Ans.  a; =49. 


63.  Given  4/5  xVx-\-2=V5x-\-2  to  find  the  value  of  a;. 

Ans.  x=-^. 

6I«  Given  ax-{-aV2ax  +  x^=:ab  to  find  the  value  of  a;. 

A  ^' 

^n«.  a;= 


2(a  +  6y 
18 


194  SIMPLE  EQUATIONS, 

Vx  +  2a_Vx-\-4:a 
Vx  +  b       Vx-hSb 


65*  Given  -= :=-— to  find  the  value  of  x. 


Ans.  X: 


(s) 


66«  Given  \/4:a+x=z2Vb+x^Vx  to  find  the  value  of  x. 

(b-ay 


Ans,  X: 


2a— b 


re rtj'       \/ X  1 

67.  Given  — zr-= —  to  find  the  value  of  x.         Ans.  x: 


J^X 


X  1— a 


68.  Given = — = to  find  the  value  of  ar. 

Vax+b      SVax  +  Bb 

Ans.  x=z — , 
a 


>,  Given  — —— -=9  to  find  the  value  of  ic.      Ans.  x=^, 

V4:X+l  —  2Vx 


70.  Given  ^«  +  ^+^^L-l::^&  to  find  the  value  of  x. 

Va  +  x—Va—x 

2ab 
Ans.  x=.^^. 

71.  Given  ^ '^  ^ i/'^^—l=2  to  find  the  value  of  a;.     Ans.  x=l^, 

x—1^  x-\-l 

72.  Given  V^6^-2^4V^6^-9  ^^  ^^  ^^^  ^^^^  ^f  ^^    ^^^  ^^^^ 

V6.r  +  2     4V^6a;+6 


73.  Given  a  +  6Va?+rf=#to  find  the  value  of  a;, 

Ans.  x=l-^\  —d, 

Ofi 

74.  Given  4/3^+ i/rr — Q = to  find  the  value  of  ar. 

♦/a;— 9 

Ans.  a:=25. 

75.  Given  a/\+W^'+\2  —  \  +a;  to  find  the  value  of  a;. 

Ans.  a;=2, 

76.  Given  i/ar  +  V'^— |/a;— V5=-( =1    to  find  the  value  of  a?. 

Ans.  xJ^. 


SIMPLE  EQUATIONS.  195 


77«  Given :  — = — : :  14  :  6  to  find  the  value  of  x, 

5  7 


Ans,  a; =4. 


78.  Given  5a; +t^ =9H — ;; ;:-  to  find  the  value  of  a;. 

4a;  +  3  2x  +  3 


Ans.  a;=3. 


79*  Given  \/9ar— 4  +  6  — 8  to  find  the  value  of  a?.         Ans,  a: =4. 

2Y 4x   15+  2x 

80»  Given  — - — : 2x  : :  5  :  4  to  find  the  value  of  x. 

Ans.  x=S* 

4x4- 14 
81.  Given  16a;  +  5  :       ;        : :  36a:  +  10  :  1  to  find  the  value  of  a?. 
vX-x"  31 

Ans.  x=5» 

Vx  +  Vh     a  ,  ^    ,  ,,         V        .  .    .  ./«  +  ^\'* 

Vx-Vh 


82.  Given  -r= ==t  to  find  the  value  of  x.     Ans.  x=h\ = ) . 

\/r.—\/h     0  \a—b/ 


83.  Given  — = = — = to  find  the  value  of  a;.    Ans.  x=4. 

Vx+2      fa;+40 


84t  Given  a  VbX'-c=d  Vex+fx^g  to  find  the  value  of  x. 

_      a^c—d^g 

85«  Given -=-  • to  find  the  value  of  a:. 

3     6^--3a:     3     x—2 

A  1* 

An^.  x=^. 


36 


86.  Given1/a:+Va;— 9=-z=  to  find  the  value  of  a;. 

4/a:— 9 

Ans.  a; =25. 

J,.     _,.        3a;  81a;»-9  „       3     2a:'-l     57- 3a;   ^     ^    . 

87.  Given— —7 ^, r-=3a;— -  • —   to  find 

2      (3a;— l)(a;4-3)  2       a; +  3  2 

the  value  of  a;.  ^w«.  a;=10. 


88.  Given    Ai|/a;'  +  39a;  +  374-i/a;»4-20a;  +  51  i  =|/?i^?    to 

19  I  )       '^   a;+  17 

find  the  value  of  a;.  Ans.  a; =78. 

89.  Given  a;»  +  2a;=(a;  +  a)»  to  find  the  value  of  x. 


Ans.  X: 


2(1— a)* 

90.  Given  a;''  +  2a;'4-aJ=(a;*  +  3a;)(a;— l)-f  16  to  find  the  value  of  a;. 

Ans.  a;=4. 


196  SIMPLE   EQlTATlOIfS. 

*91.  Given  — - —  :  x—5  : :  -  :  -  to  find  the  value  of  x,  Am^  x=:5, 
4  3    4 

92.  Given  ^        ^\ — ^— 3a= t 2a;  +  — t —    to    find    the 

a — 0  a  +  o  0 

a'  +  Sa'b+4a''b''—6ab*-\-2b^ 

value  of  a;.  Ans,  x= —tf::-^ — t T^n • 

26[2a'  +  (a— 6)6] 

93.  Given  t-H-i — I- -7-+^=^  to  find  the  value  of  x, 

bx    ax    fx     hx 

adfh  +  bcfh  +  bdeh  +  bdfg 


Ans.  X- 


bdfhk 


^.     _.        I0a;+17     12a;+2      5a?-4^    ^    ,   .         ,        . 

91 1  Given  — — — --= — - —  to  find  the  value  of  a;. 

18  \oX —  Id  y 

Ans.  a;=4« 

^^    ^.        2a;+8i      13a;-2      x     1x     x+\Q  ^     ^    .     .  . 

95.  Given— ^ ___-  +  _=_____  to  find  the  value 

of  a?.  Ans,  a?=4. 

--     _.        7a: +6     2ar  +  4f     a;     11a;     a;-3^    .    ,    .        .        . 

96.  Given  -28 23^  +  4=Tf  """42-  *^  ^^^  *^'  ^"^^^  ^^''- 

Ans.  x=4. 

^    ^.        6-5x       7-2a;'        l4-3a;     2a;-V-       1     ,     «    .   ,. 

97.  Given -^---^^^—^^-^^ ^+--  to  find  the 

value  of  a;.  Ans*  a;=4. 

98.  Given  -7 dc=bx—ac  to  find  the  value  of  a;. 

Ans.  .=ffcfJ-Z£). 
a"— 6'  +  oc 

^    ^.        18a;- 19     llar+ 21     9a;+ 15  ^    „   ^    . 

99.  Given  — -— \--- — r-TT-= — tt —  to  h^d  the  value  of  x, 

28  6a;4-14  14 

Ans.  a; =7. 
2a;— 3     3a;  — 1 


100.  Given =-  • to  find  the  value  of  a;. 

2  a;— 1  2     3a;— 2 


2 

Ans.  X——-, 
3 

*  There  is  a  peculiarity  in  tEis  example.    Any  quantities  whatever,  whether 
in  the  ratio  of  |  to  f  or  not,  when  substitued  for  §  and  |,  wiU  give  the 
answer.     Can  the  student  explain  the  reason  ? 


QUESTIONS  INVOLVING  SIMPLE  EQUATIOJ^S,   ETC.       197 


QUESTIONS  INVOLVING  SIMPLE  EQUATIONS 
CONTAINING  ONE  UNKNOWN  QUANTITY. 

QUESTION 

(266.)  1.  What  number  is  that,  the  double  of  which  exceeds  its 
half  by  6? 

SOLUTION. 

Let  x=  the  number. 

Then,  by  the  conditions  of  the  question,  we  must  have  the  equation 

X 

4tx—x=l2, 
8ar=  12, 
x=4,  the  number  required. 

Another  Solution, 
Let  2a?  =  the  number. 

Then,  by  the  conditions  of  the  question,  we  must  have  the  equation 
4x—x=6, 
3a; =6, 
x=2, 
2a; =4,  the  number  required. 

QUESTION 

2.  A  person  employed  4  workmen  ;  to  the  first  of  whom  he  gave 
2  dollars  more  than  to  the  second ;  to  the  second,  3  dollars  more  than 
to  the  third ;  and  to  the  third,  4  dollars  more  than  to  the  fourth. 
Their  wages  amounted  to  32  doUais.     How  much  did  each  receive  ? 

SOLUTION. 

Let  x=  the  sum  received  by  the. fourth, 
then  a; +  4=  "         "  "  third, 

x  +  1=  "         "  "  second, 

and  a; +  9=  "         "  "  first. 

By  the  conditions  of  the  question,  the  sum  of  these  must  equal  32 
dollars.     Therefore, 


198       QUESTIONS  INVOLVING  SIMPLE  EQUATIONS,   ETC. 

4a;+20=32, 
4rc=12, 
x=  3j  the  sum  received  by  the  fourth, 
x  +  4=:  1,        "  "  "  third, 

a; +  '7  =  10,         "  "  "-  second, 

and  a;+9  =  12,         "  «  "  first. 

QUESTION 

3.  What  two  numbers  are  to  each  other  as  2  to  8 ;  to  each  of 
which  if  4  be  added,  the  sums  will  be  as  6  to  7  ? 

SOLUTION. 

Let  2x  and  Sx  be  the  numbers. 

Then,  by  the  conditions  of  the  question, 

2X'\-4:3x  +  4:::5:1. 

Whence,   14aj  +  28  =  15a:  +  20, 
^       8^x, 

2a:  =16,  the  first  number. 
3a;=24,  the  second  number. 

QUESTION 

4.  A  person  being  asked  the  hour,  answered  that  it  was  between  five 
and  six  ;  and  the  hour  and  minute  hands  were  together.  What  was 
the  time  ? 

SOLUTION. 

Let  x=  the  time  past  5. 

Then,  since  the  minute-hand  goes  12  times  as  fast  as  the  hour- 
hand,  it  follows,  that  6  4- a;  is  12  times  x, 
.' .  \2x=b-\-x 
\\x=b 
x=j\  of  an  hour  =  27  minutes  16 j\  seconds,  the  time 
past  5  o'clock. 

QUESTION 

5.  What  number  is  that  to  which  if  1,  5,  and  13,  be  severally 
added,  the  first  sum  shall  be  to  the  second  as  the  second  to  the 
third? 


QUESTIONS  INVOLVING   SIMPLE   EQUATIONS,  ETC.        199 
SOLUTION. 

Let  x=  the  number  required. 

Then,  by  the  conditions  of  the  question, 

x  +  \  :a;  +  5  ::  x-\-b  :  a;+13. 
Whose  solution  gives  a;=3. 

QUESTION 

6.  A  shepherd,  in  time  of  war,  was  plundered  by  a  party  of  soldiers, 
who  took  \  of  his  flock,  and  ^^  of  a  sheep ;  another  party  took  from 
him  i  of  what  he  had  left,  and  -5  of  a  sheep  ;  then  a  third  party  took 
^  of  what  now  remained,  and  |  of  a  sheep.  After  which  he  had  but 
25  sheep  left.     How  many  had  he  at  first  ? 

SOLUTION. 

Let  x=  the  number  he  had  at  first. 

X      1 

Then  -+-=  the  number  the  first  party  took  away. 
4     4 

Which,  being  subtracted  from  x,  gives 

3a;     1 

— — -=  the  number  remaining. 
4      4 

The  second  party  took  away  \  of  these,  +  J  of  a  sheep,  which  left 

3\T~4/     3'^^   2     2' 
Of  these  the  third  party  took  one-half  +  i  of  a  sheep,  which  left 


2\2     2/ 


1         x     ^ 
2'"^4-4- 
X     3 

•••  4-4=  '' 

a--3r=100 
x—\0^. 

QUESTION 

Y.  A  man  and  his  wife  usually  drank  a  vessel  of  beer  in  12  days  ; 
but  when  the  man  was  gone,  it  lasted  the  woman  30  days.  In  how 
many  days  would  the  man  alone  drink  it  ? 

SOLUTION. 

Let  X  =  the  number  of  days  it  would  take  the  man  alone  to  drink  it. 
Then,  the  man,  in  1  day,  would  drink  -  of  it. 

X 


The  woman,  "  "  "         3V    " 

The  man  and  woman  together  "         y»^    " 


200       QUESTIONS  INVOLVING  SIMPLE  EQUATIONS,  ETC. 

...  i-=i+i 

12     x^S( 


5a:=60  4-2a: 
Sx=z60 
ic— 20,  the  time  it  would  take  the 


man  alone  to  drink  it. 


QUE  STION 

8.  A  person  engaged  to  reap  a  field  of  35  acres,  consisting  partly 
of  wheat  and  partly  of  rye.  For  every  acre  of  rye  he  received  5 
shillings  ;  and  what  he  received  for  an  acre  of  wheat,  augmented  by 
one  shilling,  is,  to  what  he  received  for  an  acre  of  rye,  as  7  to  3.  For 
his  whole  labor  he  received  260  shiUings.  What  was  the  number  of 
acres  of  each  sort  ? 

SOLUTION. 

Let  x=  the  number  of  acres  of  wheat ; 

Then  35— x=  the  number  of  acres  of  rye  ; 

Then  1 75  —  50:=  the  price  of  reaping  the  rye. 

By  the  question,  3:7  : :  5  : 1  + ,  the  price  of  reaping  an  acre  of  wheat* 

But  3:7::5:11|. 

Therefore,  the  price  of  reaping  1  acre  of  wheat  is  lOf^^^^-  shillings, 

32a; 
And  "         "         "         ic  acres      "      "      shillings. 

32a;       ^  ^ 

.-.    -—  +  175—   5a;=260, 
3 

32a;  +  525  — 15a:=780, 

I7a;=r255, 

ar=15,  the  No.  of  acres  of  wheat. 

35— ar=20,  "  ".         rye. 

QUESTION 

9.  The  hold  of  a  ship  contained  442  gallons  of  water.  This  was 
emptied  out  by  two  buckets :  the  greater  of  which,  holding  twice  as 
much  as  the  other,  was  emptied  twice  in  3  minutes ;  but  the  less^ 
three  times  in  2  minutes ;  and  the  whole  time  of  emptying  was  12 
minutes.     How  much  did  each  hold  ? 

SOLUTION. 

Let  X  =  the  number  of  gallons  the  less  held. 
Then  2a; = the  number  of  gallons  the  greater  held. 


QUESTIONS.  201 

4a?=  the  gallons  thrown  out  by  the  greater  in  3  minutes 
iex=  "  "  "  "  12        '* 

18a:=  "  "  "  less    "    12        " 

,-.    16a;+18a;=34a;=442, 

a;^=13,  the  No.  of  gallons  the  first  bucket  held. 
2a:=26,  "  "  second     "       " 

QUESTIONS. 

1.  What  number  is  that,  from  the  treble  of  which  if  18  be  sub- 
tracted, the  remainder  is  6  ?  Ans.  8. 

2.  What  number  is  that,  the  double  of  which  exceeds  f  of  its  half 
by  40?  Ans.  25. 

3.  In  fencing  the  side  of  a  field,  whose  length  was  450  yards,  two 
workmen  were  employed ;  one  of  whom  fenced  9  yards,  and  the  other 
6,  per  day.     How  many  days  did  they  work  ?  Ans.  30. 

4.  A  farmer  sold  13  bushels  of  barley,  at  a  certain  price;  and 
afterward  17  bushels,  at  the  same  rate;  and  at  the  second  time 
received  36  dimes  more  than  at  the  first.  What  was  the  price  of  a 
bushel  ?  Ans.  90  cents. 

5.  A  draper  sold  two  pieces  of  cloth,  by  one  of  which  he  lost  $6 
more  than  by  the  other  :  and  his  whole  loss  was  $5  less  than  treble 
the  less  loss.     What  were  the  losses  sustained  by  each  piece  ? 

Ans.  $11,  and  $17. 

6.  A  company  settling  their  reckoning  at  a  tavern,  pay  $8  each  ; 
but  observe,  that  if  there  had  been  4  more,  they  should  only  have 
paid  17  each  ?     How  many  were  there  ?  Ans.  28. 

7.  'I'wo  workmen  received  the  same  sum  for  their  labor  ;  but  if  one 
had  received  $15  more,  and  the  other  $9  less,  then  one  would  have 
had  just  three  times  as  much  as  the  other.  What  did  they  each  re- 
ceive ?  -  Ans.  $21  each. 

8.  What  number  is  that,  the  treble  of  which  is  as  much  above  40, 
as  its  half  is  below  51  ?  A7iS.  26. 

9.  A  person  has  a  certain  number  of  ijorses  at  a  livery  stable,  and 
3  times  as  many  at  grass.  He  keeps  15  in  constant  employment;  and 
his  whole  number  is  7  times  the  number  in  the  stable.  What  was  the 
whole  number  ?  Ans.  35. 

10.  Two  men  at  the  distance  of  150  miles  set  out  to  meet  each 
other  ;  one  goes  3  miles  while  the  other  goes  7.  How  much  of  the 
distance  does  each  travel?     Ans.  One  45,  and  the  other  105  miles. 


202  QUESTIONS. 

11.  A  person  put  out  a  certain  sum  at  interest  for  6i  years,  at  6 
per  cent,  simple  interest ;  and  found  that  if  had  put  out  the  same  sum 
for  12  years  and  9  months,  at  4  per  cent.,  he  would  have  received 
$185  more.     What  was  the  sum  put  at  interest.  Ans.  $1000. 

12.  From  two  casks  of  equal  size  are  drawn  quantities,  which  are 
in  the  proportion  of  6  to  7  ;  and  it  appears  that  if  16  gallons  less  had 
been  drawn  from  that  which  is  now  the  emptier,  only  half  as  much 
would  have  been  drawn  from  it  as  from  the  other.  How  many  gal- 
lons were  drawn  from  each  ? 

Ans.  24  gallons  from  one,  and  28  gallons  from  the  other. 

13.  Out  of  a  certain  sum,  a  man  paid  his  creditors  |96  ;  half  o» 
the  remainder  he  lent  his  friend ;  he  then  spent  i  of  what  now  re- 
mained ;  and  after  all  these  deductions  had  j\  of  his  money  left. 
How  much  had  he  at  first  ?  Ans.  $128. 

14.  Six  hundred  persons  voted  upon  a  disputed  question,  which* 
was  lost  by  a  certain  number.  The  same  number  of  persons  having 
voted  again  upon  the  same  question,  it  was  from  some  change  in  cir- 
cumstances carried  by  twice  as  many  as  it  was  before  lost  by  ;  and 
the  new  majority  was  to  the  former  one  as  8:7.  How  many 
changed  tht;ir  minds  ?  Ans.  150. 

15.  A  sportsman,  keeping  an  account  of  the  number  of  birds  he 
killed,  found  that  each  succeeding  season  he  wanted  50,  in  order  that 
the  number  killed  might  bear  the  proportion  of  3  :  2  to  the  number 
killed  in  the  preceding  year.  In  the  fourth  year  he  found  that  he  had 
killed  170  fewer  than  three  times  the  number  killed  in  the  first  year. 
How  many  did  he  kill  the  first  year  ?  Ans.  180. 

16.  Several  detachments  of  artillery  divided  a  certain  number  of 
cannon  balls.  The  first  took  72,  and  ^  of  the  remainder  ;  the  next 
144,  and  ^  of  the  remainder ;  the  third  216,  and  ^  of  the  remainder; 
the  fourth  288,  and  i  of  those  that  were  left ;  and  so  on ;  when  it 
was  found  that  the  balls  had  been  equally  divided.  What  was  the 
number  of  balls  and  detachments  ? 

Ans.  4608  balls,  and  8  detachments. 

17.  Two  persons,  A  and  B,  start  at  the  same  time  for  a  race  which 
lasted  6  minutes.  Now  after  galloping  4  minutes  at  the  same  uniform 
pace  at  which  each  started,  the  distance  between  them  is  j^-^  the 
part  of  the  whole  length  of  the  course.  They  continue  to  run  for  I 
minute  more,  at  the  same  speed  as  at  first ;  and  then  B,  who  is  last, 


QUESTIONS.  203 

quickens  the  speed  of  his  horse  20  yards  a  minute,  and  comes  in 
exactly  two  yards  before  A,  whose  horse  had  run  at  the  same  uni- 
form pace  throughout.     What  was  the  length  of  the  course  ? 

Ans.  3  miles. 

1 8.  A  packet  sailing  from  Dover  with  a  fair  wind,  arrives  at  Calais 
in  2  horn's;  and  on  its  return  the  wind  being  contrary,  it  proceeds 
6  miles  an  hour  slower  than  it  went.  Now  when  it  is  half  way 
over,  the  wind  changing,  it  sails  2  miles  an  hour  faster,  and  reaches 
Dover  sooner  than  it  would  have  done  had  the  wind  not  chano^ed,  in 
the  proportion  of  6  :  Y.  What  were  the  rates  of  sailing  and  the  dis- 
tance between  Dover  and  Calais  ? 

Ans.  On  its  return  it  sails  5  and  7  miles  an  hour,  and  the  distance 
is  22  miles. 

19.  A  farmer  has  a  stack  of  hay,  from  which  he  sells  a  quantity 
I  which  is  to  the  quantity  remaining  in  the  proportion  of  4  to  5.     He 

then  uses  15  loads,  and  finds  that  he  has  a  quantity  left  which  is  to 
the  quantity  sold  as  1  to  2.  How  many  loads  did  the  stack  contain 
at  first?  Ans.  45. 

20.  In  a  naval  engagement,  the  number  of  ships  taken  was  1  more, 
and  the  number  burnt  2  fewer  than   the   number  sunk.      Fifteen 

.  escaped,  and  the  fleet  consisted  of  8  times  the  number  sunk.     Of  how 
many  ships  did  the  fleet  consist  ?  Ans.  32. 

21.  At  the  review  of  an  army,  the  troops  were  drawn  up  in  a  solid 
mass,  40  deep,  when  there  were  just  J-  as  many  men  in  front  as 
there  were  spectators.  Had  the  depth,  however,  been  increased  by 
5,  and  the  spectators  been  drawn  up  in  the  mass  with  the  array,  the 
number  of  men  in  front  would  have  been  100  fewer  than  before.  Of 
what  number  did  the  army  consist  ?  Ans.  180000. 

22.  A  and  B  playing  at  billiards,  A  bet  5  dollars  to  4  on  every 
game,  and  found  that  after  a  certain  amount  of  games,  he  had  won  10 
dollars.  Had  B  won  one  game  more,  the  number  won  by  him  would 
have  been  to  the  number  won  by  ^1  as  3  to  4.  How  many  did  each 
win?  Ans.  A  won  20,  and  ^14. 

23.  A  besieged  garrison  had  such  a  quantity  of  bread  as  would,  if 
distributed  to  each  at  10  ounces  a  day,  last  6  weeks,  but  having  lost 
1200  men  in  a  sally,  the  governor  was  enabled  to  increase  the  allow- 
ance to  12  ounces  per  day  for  8  weeks.  What  was  the  number  of 
men  at  first  in  the  garrison  ?  Ans  3200. 


204  QUESTIONS. 

24.  During  a  panic,  there  was  a  run  on  two  bankers  A  and  B.  B 
stopped  payment  at  the  end  of  3  days,  in  consequence  of  which  the 
alarm  increased,  and  the  daily  demand  for  cash  on  A  being  trebled, 
A  failed  at  the  end  of  2  more  days.  But  if  A  and  B  had  joined,  their 
capitals,  they  might  both  have  stood  the  I'un,  r.s  it  was  at  first,  for  7 
days,  at  the  end  of  which  time  B  would  have  been  indebted  to  A 
$4000.     What  was  the  daily  demand  for  cash  at  ^'s  bank  at  first? 

Ans.  $2000. 

25.  There  are  two  numbers  in  the  proportion  of  ^  to  |,  which 
being  increased  respectively  by  6  and  5,  are  in  the  proportion  of  |  to 
^.     What  are  the  numbers  ?  Ans.  30  and  40. 

26.  A  gentleman  meeting  4  poor  persons,  distributed  60  cents 
among  them  :  to  the  second  he  gave  twice,  to  the  third  thrice,  and  to 
the  fourth  four  times  as  much  as  to  the  first.  How  much  did  he 
give  to  each  ?  »  Ans.  6,  12,  18,  and  24  cents  respectively. 

27.  A  farmer  has  two  flocks  of  sheep,  each  containing  the  same 
number.  From  one  of  these  he  sells  39,  and  from  the  other  93  ;  and 
finds  just  twice  as  many  remaining  in  one  as  in  the  other.  How 
many  did  each  flock  originally  contain  ?  Ans.  147. 

28.  Four  places  are  situated  in  the  order  of  the  four  letters  A,  By 
C,  D,  The  distance  from  ^  to  i>  is  34  miles ;  the  distance  from  A 
to  B ;  distance  from  (7  to  i> : :  2  :  3,  and  }  of  the  distance  from  A  to 
B  added  to  half  the  distance  from  C  to  i>  is  3  times  the  distance 
from  B  to  C.     What  are  the  respective  distances  ? 

Ans.  AB=12,  BC=4,  and  CD=18. 

29.  In  a  mixture  of  wine  and  cider,  half  of  the  whole  +  25  gallons 
was  wine,  and  i  of  the  whole  —5  gallons  was  cider.  How  many 
gallons  were  there  of  each  ? 

Ans.  85  gallons  of  wine,  and  35  of  cider. 

30.  A  footman  who  contracted  for  £8  a  year,  and  a  livery  suit,  was 
turned  away  at  the  end  of  7  months,  and  received  only  £2  3s.  4rf.  and 
his  livery.     What  was  its  value  ?  An^.  £6. 

31.  A  cistern  into  which  water  was  let  by  two  pipes,  A  and  B,  will 
be  filled  by  them  both  running  together  in  12  hours,  and  by  the  pipe 
A,  alone,  in  20  hours.  In  what  time  will  it  be  filled  by  the  pipe  B 
alone?  Ans.  30  hours. 

32*.  A  person  has  two  sorts  of  wine,  one  worth  20  pence  a  quart, 


QUESTIONS.  206 

and  the  other  1 2  pence ;  from  which  he  would  mix  a  quart  to  be 
worth  14  pence.    How  much  of  each  must  he  take  ? 

Ans.  He  must  take  ^  of  the  iSrst,  and  |  of  the  second. 

33.  A  hare,  50  of  her  leaps  before  a  greyhound,  takes  4  leaps  to 
the  greyhound's  3  ;  but  2  of  the  greyhound's  leaps  are  as  much  as  3 
of  the  hare's.  How  many  leaps  must  the  greyhound  take  to  catch  the 
hare?  Ans,  300. 

34.  Two  pieces  of  cloth  of  equal  goodness,  but  of  different  lengths, 
were  bought,  the  one  for  |5,  and  the  other  for  $6^.  Now,  if  the 
lengths  of  both  pieces  were  increased  by  10,  the  numbers  resulting 
would  be  in  the  proportion  of  5  to  6.  How  long  was  each  piece,  and 
what  was  the  cost  a  yard  ? 

Ans.  One  piece  was  20  yards  long,  and  the  other  26.     Cost,  25 
cts.  a  yard. 

35.  Two  persons,  A  and  B,  have  both  the  same  annual  income. 
A  lays  by  J  of  his :  but  B,  by  spending  $80  per  annum  more  than 
Af  at  the  end  of  4  years  finds  himself  $220  in  debt.  How  much  did 
each  receive  and  spend  annually  ? 

Ans.  The  annual  income  of  each  is  $125,  and  ^'s  annual  expendi- 
ture is  $100,  and  ^'s  $180. 

36.  An  egg-merchant  meeting  with  three  customers,  sells  to  the 
first  of  them  half  his  stock  and  1  egg  more ;  to  the  second  he  dis- 
poses of  half  the  remainder  and  2  eggs  more  ;  and  to  the  third  half 
of  ^hat  he  then  had  left  and  3  eggs  more  ;  and  afterward  discovers 
that  he  has  parted  with  his  whole  stock.  What  number  had  he  at 
first?  Ans.  34. 

37.  A  person  disposes  of  turkeys  at  as  many  dimes  each  as  the 
number  he  has,  and  returning  1  dime  finds  that  if  he  had  had  one 
more  to  sell  on  the  same  condition,  and  had  returned  2  dimes,  he 
would  have  received  20  dimes  more  from  his  bargain.  What  num- 
ber did  he  dispose  of?  Ans.  10. 

38.  A  gentleman  bequeaths  his  property  as  follows  :  To  his  eldest 
child  he  leaves  $1800,  and  }  of  the  rest  of  his  property ;  to  the  second, 
twice  that  sum  and  ^  of  what  then  remained  ;  to  the  third,  three  times 
the  same  sum  and  ^  of  the  remainder,  and  so  on ;  and  by  this  ar- 
rangement his  property  is  divided  equally  among  his  children.  How 
many  children  were  there,  and  what  was  the  fortune  of  each  ? 

Ans.   5,  and  $9000,  the  fortune  of  eachi 


206  QUESTIONS.     • 

89.  ^  and  B  possess  certain  sums  of  money,  such  that  if  they  gain 
%a  and  %h  respectively,  A  will  be  m  times  as  rich  as  B ;  but  if  they 
gain  %c  and  %d  respectively,  A  becomes  possessed  of  n  times  as  much 
as  B.     How  much  money  has  each  ? 

^  m  {nd — c)  —  n{i7ih — a) 


Arts. 


■  A^s  money  at  first. 

m — n 

Cnd—c)  —  {mb—a) 

^  '  =Bs  money  at  first. 


fourth  dav,  less  than  the  nrst. 

f.  67,  62;  58,  and  53  milesX-^^^^ 

flf  ^9.1000    ia  fn    ha    rlivirlAflV. 


m — W 

40.  The  crew  of  a  ship  consisted  of  her  complement  of  sailors  and 
a  number  of  soldiers.  Now  there  were  22  sailors  to  every  3  guns 
and  10  more.  Also,  the  whole  number  of  persons  was  5  times  the 
number  of  soldiers  and  guns  together.  But  after  an  engagement,  in 
which  the  slain  were  i  of  the  survivore,  there  wanted  5  to  be  13  men 
to  every  2  guns.     What  was  the  number  of  guns,  soldiers,  and  sailors  ? 

Ans.  90  guns,  6*70  sailors,  and* 5 5  soldiers. 

41.  An  express  set  out  to.  travel  140  miles  in  4  days,  but  in  conse- 
^^quence  of  the  badness  of  the  roads,  he  found  that  he  must  go  5  miles 

the  second  day,  9  the  third,  and  14  the  fourth  day,  less  than  the  first. 
How  many  miles  did  he  travel  each  day 

Ans. 

42.  The  estate  of  a  bankrupt,  valued  at  121000,  is  to  be  divided^ 
among  four  creditors  proportion  ably  to  what  is  due  them.  The  debts 
due  to  A  and  B  are  as  2:3  ;  ^'s  and  C's  claims  are  in  the  ratio  of 
4:5;  and  C"s,  and  i>'s  in  the  ratio  of  6:7.  What  sum  must  each 
receive  ?  Ans.  A  $3200,  B  $4800,  (7  $6000,  and  D  $7000. 

43.  There  are  two  towns,  A  and  B,  which  are  131  miles  distant 
from  each  other.  A  coach  sets  out  from  A  at  six  o'clock  in  the  mom* 
ing,  and  travels  at  the  rate  of  4  miles  an  hour  without  intermission, 
in  the  direct  road  toward  B.  At  2  o'clock  in  the  afternoon  of  the 
same  day,  a  coach  sets  out  from  B  to  go  to  A,  and  goes  at  the  rate 
of  5  miles  an  hour  constantly.     Where  will  they  meet  ? 

Ans.  76  miles  from  A,  and  55  from  B. 

44.  A  waterman  finds  by  experience  that  he  can  with  the  advan- 
tage of  the  common  tide  row  down  a  river  from  A  to  B^  which  is  18 
miles,  in  1  hour  and  a  half,  and  that  to  return  from  B  io  A  against 
an  equal  tide,  though  he  rows  back  along  the  shore,  where  the  stream 
is  only  |  as  strong  as  in  the  middle,  takes  him  just  2  hours  and  a 
qu'-.rter.  At  what  rate  does  the  tide  run  per  hour  in  the  middle, 
\y]iQTfi  it  is  the  strongest  ?  Ans.  2i  miles  per  hour. 


QUESTIONS.  207 

45.  The  ingredients  of  a  loaf  of  bread  are  rice,  flour,  and  water,  and 
tlie  weight  of  the  whole  is  15  lbs.  The  weight  of  the  rice  augmented 
by  5  lbs.  is  |  of  the  weight  of  the  flour,  and  the  weight  of  the  water 
is  I  of  the  weight  of  the  flour  and  the  rice  together.  What  is  the 
weight  of  each  ?  Ans.  Rice  2  lbs.,  flour  10^  lbs.,  water  2\  lbs. 

46.  Suppose  two  hands  of  a  watch,  (a)  and  (&),  were  together  on 
Sunday  noon,  and  the  motion  of  each  was  such  that  (a)  moved  round 
the  horary  cfrcle  in  one  hour,  and  {b)  in  l^-^  hour.  When  will  they 
be  together  again  for  the  first  time  ?  Ans.  6 1  hours. 

47.  The  rent  of  an  estate  this  year  is  greater  by  8  per  cent,  than  it 
was  last  year.  This  year's  rent  is  $1890.  What  was  the  rent  of  last 
year?  Ans.%\l50, 

48.  A  merchant  increases  his  capital  yearly  by  20  per  cent.,  and 
takes  from  it  every  year  llOOO  for  the  support  of  himself  and  family. 
After  he  had  carried  on  his  business,  in  this  manner,  for  three  years, 
he  finds,  after  deducting  the  usual  sum,  $1000,  that  his  capital  has 
increased  $200  more  than  |  of  the  original  sum.  What  was  the 
original  capital  ?  ^ws.  $30000. 

49.  A  person  has  4  wine  casks  of  different  sizes.  When  he  fills 
the  2d  empty  cask  from  the  first  full  one,  there  remains  in  the  first 
only  I  of  the  wine  ;  when  he  fills  the  3d  empty  cask  from  the  2d  full 
one,  there  is  left  in  the  2d  only  |  of  the  wine ;  but  when  he  attempts 
to  fill  the  4th  empty  cask  from  the  3d  full  one,  then  only  -^-^  of  the 
4th  is  filled,  and  if  he  wished  to  fill  the  3d  and  4th  empty  casks  from 
the  first  full  one,  then  these  would  not  only  be  filled,  but^  he  would 
have  1 5  gallons  remaining.  How  many  gallons  does  each  of  these  four 
casks  contain  ? 

Ans.  The  1st,  140 ;  the  2d,  60 ;  the  3d,  45  ;    and  the  4th,  80 
gallons. 

50.  A  father  leaves  a  number  of  children,  and  a  certain  sum  of 
money,  which  they  are  to  divide  among  them  as  follows  :  The  first 
is  to  receive  $100,  and  a  10th  part  of  the  remainder;  after  this,  the 
second  receives  $200,  and  a  10th  part  of  the  residue  ;  again,  the  third 
receives  $300,  and  a  10th  part  of  the  remainder ;  and  so  on.  At  last 
it  is  found  that  all  the  children  have  received  the  same  sum.  What 
was  the  fortune  left,  and  how  many  children  were  there  ? 

Ans.  $8100,  and  9  children. 

61.  What  number  is  that  which,  if  it  be  increased  by  7,  the  square 


208  QtTESTlONS. 

root  of  the  sum  shall  be  equal  to  the  square  root  of  the  number  itselt 
and  1  more  ?  Ans.  9. 

62.  A  person  wishes  to  give  3  cents  a-piece  to  some  beggars,  but 
finds  he  has  not  money  enough  by  8  cents ;  but  if  he  gives  them  2 
cents  a-piece,  he  will  have  3  cents  remaining.  What  is  the  number 
of  beggars?  Ans.  11. 

53.  What  ai-e  the  two  parts  of  60,  such  that  their  product  is  equal 
to  three  times  the  square  of  the  less  ?  Ans,  J5  and  45. 

54.  In  the  composition  of  a  quantity  of  gunpowder,  the  nitre  was 
10  lbs.  more  than  |  of  the  whole,  the  sulphur  4^  lbs.  less  than  }  oi 
the  whole,  and  the  charcoal  2  lbs.  less  than  |  of  the  nitre.  What 
was  the  amount  of  gunpowder  ?  Ans.  69  lbs. 

55.  A  person  engaged  to  work  a  days  on  these  conditions  :  For 
each  day  he  worked  he  was  to  receive  b  cents,  for  each  day  he  was 
idle  he  was  to  forfeit  c  cents.  At  the  end  of  a  days  he  received  d 
cents.     How  many  days  was  he  idle  ?  a     ^ 

/I'-i?. days. 

56.  What  must  the  fortune  and  number  of  children  be,  when,  in 
general,  the  first  receives  a  dollars,  together  with  the  nth  part  of  the 
remainder;  and  each  succeeding  child  a  dollars  more,  together  with 
the  nth  part  of  the  remainder,  and  it  is  found,  at  last,  that  they  have 
all  received  the  same  sum  ? 

Ans.  The  fortune,  =(71— 1)V,  and  children,  =n  —  l. 

5*7.  A  person  wishes  to  make  the  following  payments  at  4  different 

periods ;  one  sum  a  in  /,  a  sum  6  in  m,  a  sum  c  in  n,  and  a  sum  d  in 

p  months.     K  he  wishes  to  pay  his  whole  debt,  a-hb  +  c-^d,  at  once, 

at  what  period  must  he  do  it  ?  ,     , 

Ans.  ,      ^-\JL  months. 

a  +  b  +  c-^d 

68.  A  master  mason  has  engaged  a  number  of  masons  for  the 
erection  of  a  building.  He  finds,  after  entering  into  a  calculation, 
that  if  he  gave  each  man  m  shillings  a  day,  he  would  daily  expend 
a  shillings  less  than  was  assigned  for  that  purpose  by  the  estimate, 
and  that  he  would  lose  b  shillings,  if  he  should  give  each  n  shillings. 
How  many  men  did  he  hire,  and  what  was  the  daily  wages  of  each  ? 

a-\-b 


Ans.   ■< 


The  number  of  masons  was 


n—m 


The  daily  wages  of  each  was  ~  shillings. 


SmULTANEOtJS  EQUATIONS,  ETC.  209 

59.  What  number  must  be  added  to  each  of  the  two  given  num- 
bers, a  and  6,  that  the  sums  may  be  as  ??i :  w  ? 

.       Tfib — nd 

Ans.  . 

n—m 

60.  Three  masons  are  employed  in  building  a  wall.  The  first  builds 
8  cubic  feet  in  5  days ;  the  2d,  9  cubic  feet  in  4  days ;  and  the  3d, 
10  cubic  feet  in  6  days.  How  much  time  will  these  masons  need, 
when  they  work  together,  to  build  756  cubic  feet  of  the  wall  ? 

Ans.  ISTJ/y  days. 


i^  .  ■  ♦  > .  ^ 


SIMULTANEOUS    EQUATIONS    OF    THE 

FIRST   DEGREE,    CONTAINING  TWO 

UNKNOWN    QUANTITIES. 

(267.)  Simultaneous  Equations  are  such  as  must  exist  at  the 
same  time,  the  values  c£  the  unknown  quantities  in  each  equation 
being  restricted  by  the  other. 

PROBLEM. 
(268.)  To  discuss  the  nature  of  simultaneous  equations. 

DISCUSSION. 

Let  ic  +  y=:6.  This  equation  can  be  satisfied  by  the  following 
positive  integral  values  for  x  and  y. 

x~\  and  y=6, 
x=2     "     2/=4, 
a;=3     "     y=3, 
x=4:     "     y=2, 
or,  x=b     "     y—1. 
Also,  if  a;=0,  y  must  =  6,  and  if  a;=6,  y  must  =  0.     We  might 
assign  negative  values  for  a;,  and  the  corresponding  values  of  y  would 
be  obtained  by  subtracting  the  value  of  x  from  6.     Thus,  a;=— 2, 
y=6-(-2)  =  8. 

We  can  also  assign,  for  the  value  of  x,  any  proper  or  improper 
fraction,  and  the  corresponding  value  of  y  must  be  this  value  sub- 

14 


210  ELIMINATION. 

tracted  from  6.  There  are,  then,  an  infinite  number  of  values  which 
may  be  assigned  to  x  and  y  which  will  satisfy  the  given  equation. 

Suppose,  now,  we  have  ^x  +  2y—\4:.  In  this  equation,  also,  x  and 
y  may  have  an  infinite  number  of  values.  But,  if  we  'wish  x-\-y=Q 
and  3a:  +  2y=14  to  exist  at  the  same  time,  or,  in  other  words,  that 
each  of  these  equations  must  be  restricted  by  the  other,  x  and  y  can 
only  have  those  values  which  will  at  the  same  time  satisfy  both  equa- 
tions. 

"We  have,  then,  this  problem,  to  find  what  values  of  x  and  y  will 
satisfy  a; -fy = 6,  provided  3a;  +  22/ = 1 4. 

If  a; =2  and  y=:4,  the  first  equation  will  be  satisfied,  and  these  values 
will  also  be  found  to  satisfy  the  second. 

It  will  hereafter  be  found,  that  of  all  values  of  x  and  y  which  will 
satisfy  the  first  equation,  a* =2  and  y=4  are  the  only  ones  that  will, 
at  the  same  time,  satisfy  the  second  equation. 

rc  +  y=6  is  called  an  indeterminate  equation  when  it  has  no  other 
equation  to  limit  it.  In  like  manner  x  +  y-\-z:=^  is  an  indeterminate 
equation,  when  it  has  no  other  equation  to  limit  it.  In  general,  that 
equations  may  he  determinate  there  must  be  as  many  equations  as 
there  are  unknown  quantities. 

It  must  be  carefully  observed  that  the  equations  must  all  be  inde- 
pendent, that  is,  that  no  equation  be  the  result  of  an  addition,  sub- 
traction, multiplication,  or  division  performed  upon  one  of  the  others. 
Thus  a;  +  3/ = 6,  and  a;  +  y  +  3  :=  9  are  not  independent  equations.  The 
same  may  be  said  of  x-{-y=Q,  and  a^  +  y— 3=3 ;  a;  +  y=6  and 
2x  +  2y=zl2;  x  +  y=6  and  ^x  +  lyz=z2. 


ELIMINATION. 


(269.)  Elimination  is  the  process  of  deducing  fi-om  two  or  more 
simultaneous  equations,  a  single  equation  containing  one  unknown 
quantity. 

(270.)  Simultaneous  equations  are  of  the  first  degree  when  the 
equation  deduced  from  them  is  of  the  first  degree,  of  is  a  simple 
v^quation. 


ELIMINATION.  211 


ELIMINATION    BY    SUBSTITUTION. 

(27  1  •)  Mimination  by  substitution  is  finding  an  expression  for 
the  value  of  an  unknown  quantity  in  one  equation,  and  substituting 
this  value  for  the  same  unknown  quantity  in  another  equation. 

PROBLEM. 
(272.)  To    eliminate    by   substitution   from    two    simultaneous 
equations. 

RULE. 
J^ind  an  expression  for  the  value  of  an  unknown  quantity  in  one 
equation,  and  substitute  this  expression  for  the  same  unknown  quantity 
in  the  other  equation,  and  there  will  result  a  single  equation  contain- 
ing  but  OTie  unknown  quantity. 

PROBLEM 

(273.)  1.  Given  \  ""^  +^2/=^  (1)  )  ^  ^^  ^^^  ^^^^^^  ^^  ^  ^^ 
^  ^  f  mx-\-ny=d  (2)  ) 

SOLUTION. 

y= — 7—  (3)=  value  of  y  in  (1). 

nc—nax  ,  ,      ,  . 

nyz=: —  (4)  =  (3)xw. 

nc—nax      ,  /  x       (  value  of  ny  in  (4)  sub- 

"'^+— r-^"^  <^)=istitutedin(2). 

hmx+nc  —nax=bd. 

hmx — nax  =:bd — nc. 

(bm—na)x=bd—nc. 

bd—nc 


bm—na 


(8). 


abd—acn  ..      .. 

ax—— ~  (9)  =  (8)  X  a. 

om — na  '      ^  ' 


abd—nac      i  _  /-iA\-_i     value  of  ax  in  (9) 

bm—na  ^     ^~~{  substituted  in  (1). 


,  abd—nac 

by=c . 

bm—na 

c     abd — nac 

b     b'^m—abn 

cm— ad 

bm—na 


212  ELIMINATION. 

PROBLEM 


(  4x4-   9y=51  (1)  ) 
2.  Given   j  g^_ ^g  _  q  /o(  f  *<>  ^J^d  the  values  of  x  and  y. 


SOLUTION, 


4a;=61— 9y  (3) =(1)  transposed.    . 

8a;=102-18y         (4)=(3)x2. 

102-18y-13y==9  (5)=  \  ^^lue  of  So; in  (4)  substituted 

(in  (2). 
— 3  ly  = — 93.  (6) = (5)  transposed  and  added. 

y=3.  (7)  =  (6)-^-31. 

Since,  y=3,  we  have  instead  of  4a;=51— 9y  the  equation 

4ic= 5 1  —  27 = 24    (8)  =  value  of  9y  substituted  in  (8). 
x=ze. 


PROBLEM 


3.  Given    j    ,_  3_o  Lx  f  to  find  the  values  of  x  and  y. 

SOLUTION. 

The  second  of  these  equations  is  not  of  the  first  degree,  but  the 
equation  found  by  eliminating  will  be  of  the  first  degree.  The  second 
equation  is  in  fact  the  product  of  x—y=l  by  3,  for  it  may  be  put  in 
the  form  {x-\-i/){x—y)=3,  and  we  know  that  x-\-7/=B  from  the  first 
equation. 

We  have  then  to  find  the  values  of  x  and  y  in  the  equations 
x  +  7/=3  (3). 

x-i/=l  (4). 

x=l+y  (5) = (4)  transposed. 

(6)=  value  of  a?  in  (5)  substituted  in  (8). 


2ar— y 


PROBLEM 

4-14=18(1) 


4.  Given  ^  gvlar 

^2H:^  4. 16=19  (2) 


►  to  find  the  values  of  x  and  y. 


ELIMINATION.                                         21t 

SOLUTION. 

2a?— y=8 

(3)=(1)  X  2  and  terms  transposed. 

2y+a;=9 

(4)  — (2)  X  3  and  terms  transposed. 

x=9—2i/ 

(6)— (4)  transposed. 

2a?=18— 4y 

(6)^(5)  X  2. 

18-4y-y=8 

(7)=  value  of  2x  in  (6)  substituted  in  (3). 

-6y=:-10. 

y=2. 

a;=9— 2y. 

a:=:9-4. 

a;=6. 

EXAMPLES. 

!•  Given 


2,  Given 


3«  Given 


!•  Given 


5.  Given 


•I  «       «       . «  f  to  find  the  values  of  x  and  y. 

^7J5.  fl;=3,  y=2. 

iQ^ Qt/z^   9   ) 
*,       .  ^  f  to  find  the  values  of  x  and  y. 

^ns.  a;=3,  y=l. 

■j  ^         y~  r.^  f  to  find  the  values  of  a;  and  y. 
\  15a;  +  2y=66,  i  ^ 

Ans.  ic=4,  y=3. 

»,       ^       , «   f  to  find  the  values  of  x  and  y« 

Ans,  a;=:l,  y=l. 


1 


4y  .  *lx—2y     ^ 


to  find  the  values  of  x 
and  y. 

^rw.  a;=4,  y=ll. 


6«  Given  ^ 


o  4 


to  find  the  values  of  x  and  y. 
^W5.  a;=4,  y=ll. 


1 


I7a;+lly=:40  ) 
7.  Given  i  o  _  q  i   *^  ^^^  *^®  values  of  x  and  y. 


^W5.    a:=-Tr:r,  y  = 


179 
69' 


137 
69* 


/fir  )c  *>&^  -^1^ 

214  ELIMINATIOK  BY  COMPARlSOiT. 

w       ^       ,.,'!•  to  find  the  values  of  x  and  V. 


9. 


Given  ]  o   _oo'  f  *^  ^^^  ^®  values  of  a;  and  y. 

(  *^ — 23/ — 22,  J 


ir 


^4715.  a:=4,  y=3. 
f  6a;-4- 4?/r=26   ) 

10.  Given   \  ^       / '  >•  to  find  the  values  of  x  and  y. 

(  6a;-}-4y=28,  )  ^ 

-47W.  a:=2,  y=4. 

ELIMINATION    BY    COMPARISON. 

(274.)  Elimination  by  Comparison  is  finding  an  expression 
for  the  value  of  an  unknown  quantity  in  one  equation,  and  also,  an 
expression  for  the  value  of  the  same  unknown  quantity  in  another 
equation,  and  putting  the  expressions  equal  to  each  other. 

PROBLEM. 
(275.)    To    eliminate   by  comparison   from   two    simultaneous 
equations. 

RULE. 
Mnd  an  expression  for  the  value  of  one  of  the  unknown  quantities 
in  the  first  equation,  and  put  it  equal  to  an  expression  for  the  value 
of  the  same  unknown  quantity  found  from  the  second  equation.) 

PROBLEM 

(276.)  1.  Given  I  ll'^l^Zn  (2)'  [  *""  ^""^  *^^  values  of  a;  andy. 

SOLUTION. 

ax=zm—hy  (3)=(1)  transposed. 

x=^  (4)=(3)^a. 

cx=n—dy  (S)=  (2)  transposed. 

_n'—dy 
c 
--*  m — by     n — dy 
a     ~      c 
cm — hey  :=  an — ady 
ady—hcy^=^an—cm 
{ad—bc)y=an — cm 
an— cm 
^     ad— be 
The  value  of  x  may  be  obtained  by  equating  the  expressions  for  the 


(6)=(5)-c. 

(7)=  value  of  x  in  (4)  and  in  (6)  equated. 


ELIMINATION   BY   COMPAEISON.  215 

value  of  y,  or  it  may  be  obtained  from  either  of  the  given  equations 
by  putting  in  it  instead  of  y  its  value  as  found  above.    Proceeding 

by  either  of  these  methods,  we  would  find  ^=— i — r-» 

PROBLEM 

(277.)  2.  Given  i  ^^^^l^^^^^^^  f}\  to  find  the  values  of 
X  and  y. 

8  OLUTION. 

100-3y 

^=-11- 

4+7y_lQ0— 3y 
4     ~~    11 
44  + 77^=400-- 12y 
89y=366 


4+7y 
y=4.  Since,  a;= — — ^  and  y=4, 


1,  4+28     „ 

we  nave  «= — ^ — = 8. 


PROBLEM 

8.  Given    \  ^'':^l'f^^,l  ^J^  \  to  find  the  values  of  a;  and  y. 
(  64a;— 3yVy=12  (2),  ) 

SOLUTION. 

3y^y=69—   7a;  (3)  =  (1)  transposed. 

S^y=*lx—  59  (4)  =  (3)  with  signs  changed. 

— 3yV2/=12— 64a:  (5)  =  (2)  transposed. 

7i;— 59  =  12  — 64a;  (6)=  value  of —3yLy  in  (4)  and  in  (5)  equated 

a;=l.  Since  3yVy=59  — ^a;  and  a;=l,  we  have 
3^^=69-7=52 
52y=l7.52 
y=11 

Hemark. — This  solution  shows  that  all  that  is  necessary  in  eliminating  by 
comparison,  is  to  find  an  expression  for  the  value  of  one  of  the  unknown  quan- 
tities when  affected  by  a  certain  coefScient,  and  then  find  from  the  other  equa- 
tion an  expression  for  the  same,  and  then  equate  them. 


216 


ELIMINATION  BY  COMPARISON. 


!•  Given  \  ,   r  to  find  the  values  of  x  and  y. 

(  x—y=b, ) 

a-^b        a— 6 
Ans.  «=-2-»  y^-g-- 

*2«  Given  i    '  '  {•  to  find  the  values  of  x  and  y. 

(  a;  +  6y=191,  ) 

Ans.  a;=16,  y=:36. 

3.  Given  •!       ~       !■  to  find  the  values  of  x  and  y. 

(  x-^y=zc  ) 

be              ac 
Ans.  x= — — r,  y= r. 

(       Yv^=2iC— 3t/        ) 

4.  Given  i  ,^       ^^       ^^!. ,    !■  to  find  the  values  of  a;  and  y 

(  19a;=60y  +  621i^,  ) 

Ans.  a;=88f,  y=lT|. 

^    ^.         j  lBx-\-Ty—S41=1ly-{-4SlXj  )   to  find  the  values  of  x 
^•^^^^^  j  2x  +  ^y=l,  \     andy. 

^n5.  a:=:  — 12,  y=60. 

ig^ gx^j^Yv 44  ) 
«  ^         =         f  to  find  the  values  of  x  and  y. 
2a:=y  +  4,        J 

^w».  a;=4|,  y=8^. 

7.  Given  i         «    ~«>,    f  to  find  the  values  of  a;  and  y. 
i  y  +  Sx=2lf  ) 

Ans.  a;=8,  y=3. 

{4ic4-9v=5l    ) 
,^  f  to  find  the  values  of  x  and  y. 
8a;— 13y=9,  )  ^ 

Ans.  a;=6,  y=3. 
7  + a;      2a;-- y 

to  find  the  values  of  x 
and  y. 


9.  Given 


5y— V     4a;— 3 

+  — ^r— =18-5a;. 


10.  Given 


2 

3a;— 1 

5 
3y-5 

6 


+  3y— 4=15 

+  2a;— 8='7| 


^n5.  a;=:3,  y=2. 

►  to  find  the  values  of  x  and  y. 
^ws.  a;=7,  y=z5. 


Multiply  the  2d  equation  by  3,  and  equate  the  values  of  3aj. 


ELIMINATION  BY  ADDITION  AND  SUBTRACTION.       217 


ELIMINATION    BY    ADDITION    AND    SUBTRACTION. 

(278.)  Elimination  hy  addition  and  subtraction  is  multiplying 
or  dividing  two  equations  so  as  to  make  the  coefficients  of  one  of  the 
unknown  quantities  the  same  in  both  equations,  and  then  subtracting 
or  adding  these  equations  according  as  the  signs  of  these  terras  are 
like  or  unlike. 

PROBLEM* 

(279.)  To  eliminate  by  addition  and  subtraction  from  two  simul- 
taneous equations. 

RULE. 

Multiply  or  divide^  if  necessary,  in  such  a  manner  as  to  cause  one 
of  the  unknown  quantities  to  have  the  same  coefficient  in  both  equa- 
tions ;  and  then  add  these  equations  if  the  signs  of  these  terms  are 
unlike,  or  subtract  one  from  the  other  if  the  signs  are  alike, 

PROBLEM 

(280.)  1.  Given  ]'^'^*^^'^^jHtofindthevaluesofa;andy. 
V  cx-y-dy — n  (2K  ) 


SOLUTION. 

a€x-\-bcy=cm  (3)  =  (l)xc, 

acx-]-ady=an  (4)  =  (2)xa, 

{bc—ad)y=cm—an  (5)  =  (3)  — (4), 

cm— an  .,      ,  .      ,, 

^=^^3^  (6)=(5)-(6«-a<0. 

The  value  of  x  may  be  obtained  in  the  same  way  by  multiplying 
the  first  equation  by  d,  and  the  second  by  b,  and  subtracting,  or  by 
substituting  in  either  of  the  given  equations  the  value  of  y  as  already 

found.     By  adopting  either  of  these  modes,  we  would  get  x= — - — 7--. 

Remark. — ^We  should  always  eliminate  those  terms  which  require  the  least 
preparation.  In  the  example  just  given  there  is  no  preference,  as  the  operation 
can  not  be  shortened,  because  a  and  c,  and  b  and  d  are  prime  to  each  other. 


218       ELIMINATION  BY  ADDITION  AND  SUBTEACTION. 

PROBLEM 


2.  Given 


1 


.,_?+i.  ,  +!^  (1), 


3 


2a;+l 


(2), 


to  find  the  values  of  ot 
and  y. 


SOLUTIO 

48y— l'7a;=156 
2y  +  30a:=160 
48y+'720a;=3840 
'737a:=:3685 


(3)=(l)x20,  &c, 
(4)=(2)x6,&c., 

(5)  =  (4)x24, 

(6)  =  (5)-(3), 
('7)=:(6)-'737, 


2?/ = 1 60  —  30a;         (8) = (4)  transposed. 
.2y=160— 150=10, 
y=5. 


PBOB  LEM 


8.  Given  \   ^      c.       ^^   rJ\  \  ^  ^^^  ^^  values  of  a;  and  y, 
(  6x-^y=^  (2),  3 


SOLUTION. 

16a;-10y=20  (3)  =  (2)  x  3, 


31a;=62 

a;=2 
10y=42  — 16a; 
.10y=42-32=10, 


(4)  =  (l)  +  (3), 
(5)=(4)-31, 
(6)=(1)  transposed, 


EXAMPLES 


1,  Given  - 


2t  Given 


6  +  4-^' 

?+^=5^ 
4^6     ^ 


to  find  the  values  of  x  and  y. 

^ws.  a;=12,  y=16. 


9a;  + 


:70, 


>-  to  find  the  values  of  x  and  y, 

Ans.  a;=6,  y=10. 

S.  Given  1("  +  ") (^  +  ^)^("+'^(^-'> +  "'•[  to findthe  value, 
(  2a?+10=3y+l'  ) 

of  a;  and  y.  ^^i«.  ar=3,  y=5. 


ELIMINATION  BY  ADDITION  AND  SUBTEACTION.  219 

f     a    _     b 
4«  Given    <  b  +y'^Sa-\-x   }-  to  find  the  the  values  of  x  and  y. 
[aa?+26y=c 


Ans.    x= 


26'— 6a'  +  c  Sa'—b^+c 


5*  Given 


3a 


Bi±4x  9y  +  33 

6a;— 4y  lly— 19 

y—3 —-^=x —, 

^2  4 


y= 


86 


to  find  the  values  of 
X  andy. 


Ans,  x=6,  y=6. 


f  3a;  +  4y  +  3     2a;  +  '7-y     ^  ,  y-8  ' 


6.  Given  * 


10  15  6 

9y  +  5a;— 8     x-\-y     1x-{-G 


12 


11    ' 


to  find  the  values 
of  X  and  y. 


Ans,  x=1f  y=9. 


7,  Given  - 


'6a;  +  13      By— 3a;— 5_       ^a;— 3y+l  " 
2  6  ""^"^         3         ' 


x  +  l   3y— 8 


3  4 

values  of  x  and  y. 


4a;::4:21, 


8.  Given 


^      21— Qy  3 

„       21-4y     18a;+13     „, 


to  find  the 

Ans.  x=z5t  y^4i. 

to  find  the  values  of 
X  and  y. 


^W5.  a;=7,  y=5. 


9«  Given 


r,«       .       ,      128a;''-18y'*  +  2l7  ^ 

16a;  +  6y— 1  = ^-^ , 

^  8a;— 3y  +  2       ' 


=5 


54 


10a;  +  10y— 35 
2a;  +  2y  +  3         '      3a;  +  2y— 1' 

4a;+3y  + 


to  find  the  values 
of  X  and  y. 


10.  Given 


2a;  +  4=3y-+ 


Ans.  a;=6,  y=6. 

24  +  5^y_16a;''4-12a;y— 8a;  +  5y  +  28 
2a;4-l"~  4a;-2  ' 

8a;''— 1  By'' +  108 


to 


4a;4-6y  +  3     ' 


find  the  values  of  x  and  y. 


Ans,  a;=3,  y=2. 


220 


MISCELLANEOUS  EXAMPLES  IN  ELIMINATION. 


MISCELLANEOUS  EXAMPLES  IN  ELIMINATION 

PROBLEM 
X 


(281.)  1.  Given 


-f7y=99(l) 
^  +  1x=5l(2) 


>  to  find  the  values  of  x  and  y. 


SOLUTION. 

(3)=(l)x7. 

(4)^(2)  X  7. 

(5)=[(3)  +  (4)]-f-60. 

(6)=(3)-(5). 

(7). 

(8) = value  of  y  in  (1)  substituted  in  (6). 


a;-f.49y=693 
49x-\-y=S5l 
x+y=:21 
48y=672 
y=   14 
a;  +  14=  21 
x=1 
The  above  artifice  can  always  be  adopted  when  the  coefficients  of 
X  and  y  in  the  first  equation  are  the  same  as  those  of  y  and  x  in  the 
second. 

PROBLEM 


2.  Given  ^ 

■  to  find  the  ralttes  of  tc  and  y. 

SOLUl 

ION. 

1     1_4 
X    y~21 

(8)  =(1)-U7. 

17     17__68 
X      y~2l 

(4)=(3)xl7. 

13     IS 

y~  "^ 

(6)=(2)-(4) 

y=1 

(«)• 

1__4      1 
x'^2\'^y 

(7)=(3)  trans. 

14      11 
•*•  x~2l'^1~3' 

x=S, 

EX  A  MI 

LBS. 

1,  Given  h 

the  values  of  x  and  y. 

Ans.  x=e.  «=:4 

MISCELLANEOUS   EXAMPLES  IN  ELIMINATION.  221 

it  Given  •}  ^       ,       ^'  r  to  find  the  values  of  x  and  y. 

(  3a;  +  iy=29,  ) 

Ans,  a;=9,  y=6, 
3.  Given   \  2^+3^=14,  )   ^^  ^^  ^^  ^^^^^  ^^  ^  ^^^ 

^ri«.  aj=24,  y=6. 

!•  Given  ■!  ®       ,    ~.„.'  r  to  find  the  values  of  x  and  y. 
(  8a;H-iy=131, ) 

Ans,  a:=16,  y=24. 

(  4ic4-v=34  ) 
5«  Given  •<        /      , «  f  to  find  the  values  of  x  and  y. 
(a;  +  4y=16,  ) 

^7w.  a; =8,  y=2. 

6«  Given  i  w        T  r  to  find  the  values  of  x  and  y. 

C  — a;  +  '7y=33j 

^W5.  x=z2,  y=^6. 

{a?4-y=8       ) 
o      o     !.  „  f  to  find  the  values  of  x  and  y. 
a;'— y'=16, ) 

Ans,  x=5,  y=8. 

8.  Given  \    •      ,    \  }•  to  find  the  values  of  x  and  y. 
(  x^—y^—h, ) 


a'  +  J    _a}—b 
2a  *^""~2a' 


^IW.  x•=—:r—^y^=' 


9«  Given 


10.  Given 


2^ 
3 


+  6y=23, 


Vy 

4 


h  to  find  the  values  of  x  and  y. 

JItw.  a;=— 3,  y=6. 


^-12=1  +  8, 
__4-__8_-^  +  27, 


to  find  the  values  of  a: 
and  y. 

^7i«.  a;=:60,  y=40. 


11,  Given  i  2   +  32/     8,      )     ^  ^^^  ^j^^  ^^^^^  ^^  ^  ^^^ 

Ans,  x=6,  y=16. 

4a;     5y     9     ,    ] 

I-— = 1 

,  x^      y^     y 

12.  Given  '(    ^       ^      h     «  r  t<^  fi^d  t^®  values  of  a?  and  y. 

-  +  -=-  +  -, 

a;       y      a?     2'J  Ans,  x=z4,  y=2. 


I 


222         MISCELLANEOUS  EXAMPLES  IN  ELIMINATION. 


13*  Given  - 


2y— a;     ^^     69— 2a; 

x—-^ =20 — , 

23— a;  2      ' 


to  find  the  values  of  x 


J^^^     ^^3-3y  f   andy. 
^"^a;-18  3 


Ans,  a;=21,  y=20. 


f  3a;  +  2y     5a;— |y+  1  _  y— 2a;_  4a;— y  ^ 

14.  Given  <5  3  ""^"^10  Y'V 

(  y  +  2a; :  y— 2a; : :  12a;  +  6y— 3  :  Qy—  12a;— 1,  ) 


to  find  the  values  of  x  and  y. 

16+60a;     16a;y— lOt 


15*  Given  ^ 


8a; 


3y-l  5  +  2y     ' 

«     «        «       2'7a;»--12y*4-38 


^W5.  a;=l,  y=4. 


to  find  the  value 
of  a;  and  y. 


^7W.  a?=2,  y=3. 


16i  Given  •< 


+  6y  +   1 


6a;'+130— 24y'  ' 
2a;— 4y  +  3     ' 


3a;- 


151  — 16a;_9a;y— 110 
4y-l    "^    3y-4   ' 

4a;*-y(a;  +  3y) 


to  find  the  valuei 
of  X  and  y . 


(3y  +  ll="^-y^^  +  ^^^4-31-4., 
17.  Given   ^    ^  a;-y  +  4  ' 

(  (a;+'7)(y-2)  +  3  =  2a;y-(y-l)(a;  +  l), 


values  of  x  and  y. 
18.  Given 


Ans,  a;=9,  y=2. 

to  find  th< 
An9,  xz=4c,  y=3. 


(    ^y—Yy  —  X=V20  —  X, 

I  Vy^ :  |/20— a; : :  3  :  2, 


to  find  the  values  of  a;  and  y, 
Ans,  a;=16,  y=26. 

•-.    ^.         (168H-19a;4-fVy=12fa;4-1084, )    ^    .    ,  ^,        , 
19.  Given  ]  ,,  J J,9^^3i;^_  jy,  [  to  find  the  valu. 

of  a;  and  y.  Ans.  a;=— 2*71,  y=136J. 


20.  Given 


3a;  +  5y 
ab^c 


{Sa—2b)ab 
''     a'-b'     ' 


to  fin( 


a'x z  +  (a  +  6+c)6y=&'a;  +  (a+26)a6, 

Or  •J-  0 


the  values  of  x  and  y. 


-4»«.  a;= — r^  y 


ab         _  ab 
a—h'  ^"a  +  b' 


MISCELLANEOUS  EXAMPLES  IN  ELIMINATION.  223 


(  hcx=^cy—2b, 
21,  Given   I  ,,       a(c^—b^)     26\    ,    [  to  find  the  values  of  a;  and  y. 


6c 


a  a +  26 

Am.  ^— ,  y= , 

be  c 


22»  Given  -j 


a     b 

— I — =m. 

X    y 

c    d 

X     y 


>■  to  find  the  values  of  x  and  y. 


be— ad  be— ad 

Ans.  x=  — z,  y: 


nb—md'        me— ma 


23,  Given 


W9_^3^j^^^^     3^+4  ^ 

4         4a;— 6         *         2     '  I   to  find  the  values  of 
8y  +  7     6a;-3y_         4y-9   j   x  and  y. 
10    "^2y-8  "*     "^      6     'J 

^»s.  a; =7,  y=9. 


24.  Given  ^  ^       f  ^      ^  ^ 

U-2y— ^-^— =Y(^+y)-3(a;-yj, 

to  find  the  values  of  a?  and  y.  Ans,  x=-^\  y=A 

^  .    11 

25*  Given  - 


12y ^-I—  — £- +  i0a;  +  13, 


4  + 


11 


X  11 

— =y+ 


3a; 


2a;     3a;— 5_4a;y  +  ^5^ 
IT''  y  +  1~  6y+2V  ' 
the  values  of  a;  and  y. 


to   find 


^5.  arzrT,  y=2. 


to  find  the  values 
of  X  and  y. 


7a;     ^       3y  +  6     3a;— 2 
4        ^         5  10    _g    yg 

26.  Given  -j       5      ~"  8  "16' 

l2^3^^^*2     s^''^    ^      ''i 

Ans.  x=4:,  y=3. 

4a;— 2y  +  3     18— a;  +  5y_a;     y     1 
3  ^  - 


27.  Given  ^ 
to  find  the  values  of  x  and  y. 


.____       7yV, 


X    V     S    V    X      1 
2.-y+15:y-2.+15::^  +  -:|--+-, 

J[w«.  a;=18,  y=24. 


224 


SIMULTANEOUS  EQUATIONS,  ETC. 


^  V  11a? 

28.  Giv«n  ^  —-J^-— -^=-— ^- » 

7y  24  6  42  56y 

12a;— 15y  +  -V-:  lOy— 8a;4- V  •  =  93  — 9a; :  Qx—i^^ 
to  find  the  values  of  a;  and  y.  Ans.  a; =9,  y='7. 


29.  Given  < 


3a; 


-5y     2a;-8y-9_y 

3  12         "12"^='"*"*' 


^4.|  +  li:4a;-|-24::3i:3i 


to  find  the  values 
of  X  and  y. 


Ans,  a;=7,  y=4. 


f.  Given  ^ 


3y— 2+a;  15a;  +  -J^ 

o 


1+- 


11  ■        33      ' 

3a;  +  2y     y— 5     lla;4-152     3y  +  l 


6 


12 


2     ' 


to  find  the  values 
of  X  and  y. 

a;=8. 


i#  ..  ♦  ..  »* 


SIMULTANEOUS  EQUATIONS  OF  THE  FIRST  DEGREE 

CONTAINING  THREE  OR  MORE  UNKNOWN 

QUANTITIES. 

PROBLEM. 
(282.)  To  eliminate  from  three  or  more  simultaneous  equations. 


SOLUTION. 

Let  (A)^  (B),  and  (C),  represent  three  simple  equations,  each  of 
which  contains  three  unknown  quantities,  as  a;,  y,  and  s.  Suppose  it 
most  convenient  first  to  eliminate  z.  Then,  according  to  one  of  the 
preceding  methods,  eliminate  z  from  (A)  and  (B),  and  there  will 
result  an  equation  which  will  contain  only  x  and  y  as  unknown  quan- 
tities, which  equation  designate  by  (D).  Next,  eliminate  z  from  [A) 
and  ((7),  or  from  (B)  and  ((7),  as  may  be  most  convenient,  and  there 
will  result  another  equation,  which  also  will  contain  only  x  and  y  as 
unknown  quantities.     Call  this  equation  (IH).    We  have  now  nothing 


SIMULTANEOUS  EQUATIONS,   ETC. 


226 


more  to  do  with  (A),  (-B),  and  (C7),  but  must  operate  on  the  result- 
ing equations,  (D)  and  (U).  Now,  let  us  eliminate  7/  from  (D)  and 
{jS)j  and  there  will  result  an  equation  (F)  which  will  contain  only  x. 
Q,  K  F. 

PROBLEM 

2a?  +  3y+    52—23  (1),) 

3a;+2y+   1z=28  (2),  C  to  find  the  values  of 

5x  +  1y  +  Uz=52  (3),) 

SOLUTI ON. 

(4)=(l)x7. 
(5)=(2)X5. 


(283.)  1.  Given 

ic,  y,  and  z. 


14.r  +  21y  +  352: 
15a;  +  10y  +  352: 


161 
140 


or-lly 
22a;  +  33y  +  552: 
25a;  +  35y+650: 


-21 
253 
260 


(6)  =  (5)-(4). 
(7)=(l)xll. 
(8)  =  (3)x5. 


3a;  +   2y 
3a;— 33y 


1 
63 


(9)=(8)-(7). 
(10)  =  (6)x3. 


35y  == 

y  = 

3a;  = 
.-.     3a;  = 

YO 
2 

1-21/ 
7-4=2 

(11). 

(12)  = 

(13)  = 
1  (14)= 

=  (9)-(10). 

=  (11)^35. 

=  (9)  transposed. 

= value  of  y  substituted  in  (13). 

X   = 
J  +  6+52= 

1 
23 

(16)  = 
(16)  = 

=  (U)-3. 

=  values  of  a;  and  y  substituted  in  (1) 

52  = 

15 

* 

2  = 

3. 

•  PROBLEM 

C     ax-{-hy  -{-cz  =c?,    ) 
2.  Given  <  a'x  +  h'y  -\-c'z  =c?',  V  to  find  the  values  of  a;,y,and2. 
ia"x^h"y  +  c"z=d'\) 


SOLUTION 


Eliminating  as  directed  in  Problem  (282),  we  shall  obtain,  after 
arranging  the  terms  in  the  separate  results, 

dh'c"  +  d'h"c  +  d"bc'-d'hc"-'dh"c'-d"h'c 


x= 


z= 


ab'c"  +  a'b"c+  a"hc'-a'hc"-ah"c'  '--a"h'c 
ad'c"  -\-a'd"c  +  a"dc'  -a'dc"  -ad"c'  -a"d'c 
'ab'c"  +a'b"c  -{-a" be' -a' be"  —ab"c'—a"Vc 
ab'd"  +  a!b"d  +  a"bd'-a'bd"-ab"d'-a"b'd 


ab'e"  +  a'b"e  +a"be'  -  a'bc"  -  ab"c'  -  a"b'c 
15 


226  SIMULTANEOUS  EQUATIONS,   ETC. 

If  tlie  student  can  only  recollect  these  general  values,  he  need  only 
make  the  proper  substitutions  without  being  compelled  to  go  through 
the  process  of  elimination. 

The  denominators  are  all  alike,  and  do  not  contain  any  of  the  ab- 
solute terms.     This  denominator  can  be  formed  as  follows : 

Write  the  coefficients  of  the  unknown  quantities  as  in  the  problem, 
repeating  the  first  two  rows ;  thus. 


a 

b 

c 

a' 

V 

c' 

a" 

b" 

c" 

a 

b 

c 

a! 

b' 

c' 

Omitting           ,  at  the  right  hand 

con 

\  hand  comer  below,  we  have, 

1st.  a 

2d.  a' 

b' 

3d.  a" 

b" 

c" 

b 

c 
c' 

,,  at  the 


The  product  of  the  terms  in  each  of  these  lines  will  give  the  posi- 
tive terms  of  the  common  denominator. 

Again,  omitting    ,         at  the  left  hand  comer  above,  and  ,,        , 

at  the  right  hand  comer  below,  we  have 


b' 

e 

6th.  a" 

b" 

c 

5th.  a 

b 

4th.  a' 

The  product  of  the  terms  in  each  of  these  lines  will  give  the  negative 
terms  of  the  common  denominator. 

We  can  write  the  respective  numerators  from  the  common  denom- 
inator by  changing  a  into  d^  a'  into  d\  and  a"  into  d"  for  the  nu- 
merator in  the  value  of  a; ;  6  into  <?,  b'  into  d\  and  b"  into  d"  for  the 
numerator  in  the  value  of  y  ;  and  c  into  o?,  c'  into  d\  and  c"  into  d" 
for  the  numerator  in  the  value  of  z. 

It  is  not  necessary  that  the  student  should  in  practice  make  the 
oraissions  which  we  did  above,  for  we  can  get  from 


SIMULTANEOUS  EQUATIONS,   ETC.  227 


a 

b 

c 

a' 

h' 

c' 

a" 

h" 

c" 

a 

h 

c 

a' 

h' 

c' 

by  passing  downward  to  the.  right,  the  terms  ab'c'\  a'b"Oj  a"hc\  and 
\>j  passing  upward  to  the  right,  the  terms  a'hc'\  ah"c\  a"h'e. 

In  the  same  manner,  by  putting  d  for  o,  d'  for  a\  and  d"  for  a"  we 
should  have  from  d        h        c 


d'  h'  c! 
d"  b"  c" 
d        b        c 

d'       V       c' 
by  passing  downward  to  the  right,  db'c'\  d'b"c,  and  d"bc'  for  the 
positive  terms  in  the  numerator  of  the  value  of  a?,  and  by  passing  up- 
ward to  the  right,  d'bc'\  db"c\  and  d%'c  for  the  negative  terms  in 
the  same  numerator. 

In  the  same  way  from     a       d       c 

a'       d'       c' 

a"      d"     c"     . 

a        d        c 

a'       d'       c' 
we  can  get  the  numerators  in  the  value  of  y. 
Also,  from  a        b        d 

a'       b'       d' 

a"      b"      d" 

a        b        d 

a'       V       d! 
we  can  get  the  numerator  in  the  value  oiz 

PROBLEM 

C  3a;-|-2y4-52=59,  ^ 
8.  Given  <  4a;+  y  +  3s=41,  V   to  find  the  values  of  a;,  y,  and  z, 

(Sar  +  Vy-f  2^  =  75,  ) 

SOLUTION.  . 

From  3  2    6  ^: 

4  1     3 

8  '/     2 

3  2     5 

4  13 


228 


SIMULTANEOUS  EQUATIONS,   ETC. 


we  get  3x1  x2  +  4x7x5-f8x2x3— 4x2x2 
X  5  for  the  common  denominator  of  a;,  y,  and  z. 
From  69     2     5 

41  1  3 
75  1  2 
69  2  5 
41  1  3 
we  get  for  the  numerator  in  the  value  of  a;, 
69x1x2+41x7x5  +  75x2x3—41x2x2 


3x7x3—8x1 


59x7x3—75x1x6. 


These  expressions  simplified,  give  «= 


2003  —  1778     225 


194  —  119 


-^=^- 


From 


3 

69 

6 

4 

41 

3 

8 

76 

2 

3 

69 

5 

4 

41 

3 

We  get  for  the  numerator  in  the  value  of  y, 
3x41x2+4x75x5+8x59x3—4x59x2—3x76x3—8x41x5=3162-2787. 

^,       ^  3162-2787     375      ^ 

Inereiore,  y. 

Also,  from 


194-119 

75-^ 

3 

2     69 

4 

1     41 

8 

7     76 

3 

2     69 

4 

1     41 

We  get  for  the  numerator  in  the  value  of  z 

3x1x75+4x7x59  +  8x2x41-4x2x75-3x7x41-8x1x59=2633-1933. 

2533  —  1933     600      ^ 
Inererore,  g=   _^  . — rT^-=-;rTr=8. 


194  —  119 


75 


PROBLEM 

C  3a;— 6^+0=- 32,  \ 
4.  Given    ^  2a:+8y      =     16,  >   to  find  the  value  of  a;,  y  and  z, 
i  5x-\-1y—2z=z     6,  ) 


SOLUTION. 

3  -6 

1" 

—  32  -6       1' 

3  —32 

1' 

3 

-6  —32 

2       8 

0 

16       8      0 

2       16 

0 

2 

8       16 

5      7 

-2 

-,       5       7-2 

-,5         5- 

-2 

- , and  5 

7        6 

3  -6 

1 

—32  —6       1 

3  —32 

1 

3 

-6-32 

2       8 

0- 

16       8      0. 

2       16 

0- 

2 

8       16 

^ 


y= 


SIMULTAISIEOUS  EQUATIONS,    ETC.  229 

We  get  by  proceeding  as  before 

(-32x8x-2+16x7xl+5x-6x0)-(16x-6x-2+-32x7x0+5x8xl) 

~(3x8x-2+2x7xl  +  5x-6x0)-(2x-6x-2  +  3x7x  0  +  5x8x1) 

_624  — 232_   392_ 

—  _34_64~— 98~~   * 

(3  X 16  X  -2+2  X  5  X 1+5  x  -32  x  0)-(2  x  -32  x  -2+3  x  5  x  0+5  x  16  k  1) 

_ 

—  86  —  208      —294     ^ 


D  -98 

(3x8x5+2x7x— 32+5x— 6x16)— (2x— 6x5+3x7xl6+5x8x— 32) 


—  808  +  1004 


196 


6.  Given 


D  ~— 98 

PROBLEM 

U+y+2=3i,  (1)) 

<  a;+y— 2=25,  (2)  >•   to  find  the  values  of  a;,  y  and  z, 

(  x—y—z=  9,  (3)  ) 


6.  Given 


SOLUTIO  N. 

2a;=40  (4)=(l)  +  (3)' 

a;=20  (6). 

2^=  6  (6)=(l)-.(2). 

.=  3  (7). 

2y=16  (8)=(2)~(3). 

y=  8. 

PROBLEM 

2a;— 3y  +  20=13,  (1) 
42;--2ar=30,  (2) 
4y  +  22!=14,  (3) 
5y  +  3i;=32,  (4)J 

SOLUTION. 

1y-2x=  .  1  (5)  =  (3)-(l) 

12v  — 6a?=  90  (6)  =  (2)x3. 

20y  +  12t;rr:128  (7)  =  (4)x4. 


to  find  the  values  of  v,  «,  y, 
and  z. 


20y  +6a:=-  38 
21y  — 6ar=     3 


(8)  =  (7)--(6). 

(9)  =  (5)X3. 


41y=  41  (10)  =  (8) +  (9). 

y=     1  (11). 

2a;=:7y— 1  (12) =(5)  transposed. 

2x=  7-1=6  (13). 

a;=  3  (14).  [forward 


230 


SIMULTANEOUS  EQUATIONS,   ETC. 


(2)  transposed. 


36 


(15): 

(16). 

(l7)  =  (16)-4. 

(18) =(3)  transposed. 


4:V=  6  +  30: 
v=  9 
22=14—4?/ 
22=14-4=10. 

2  =  5. 

Note. — The  student  should  accustom  himself  to  apply  the  general 
formula  which  has  been  given,  so  that  it  may  be  called  up  at  any 
time.  It  will  often  save  much  labor,  as  will  be  proved  to  be  the  case 
in  some  of  the  following  examples.  There  are  many  curious  proper- 
ties which  belong  to  the  general  formula  for  the  values  of  the  un- 
known quantities  as  deduced  from  three  simultaneous  equations.  But 
we  leave  them  to  be  discovered  by  the  student. 

EXAMPLES. 

rx+y  +z    =29,^ 


1.  Given 


2.  Given 


3*  Given   <az-{-cx 


(ay  +  6a?=c, 


to  find  the  values  of  a?,  y,  and  z. 
Ans.  x=8,  y=9,  2=12. 

to  find  the  values  of  x,  y,  and  2. 
Ans,  x=l,  y=2,  2=8. 


to  find  the  values  of  x,  y,  and  2. 


Ans. 


x=- 


2  =  - 


25c 

2ac 
a'  +  b'-c' 


4t  Given 


ax-\-b7/=:a' 


cx  +  dv=:  c', 
^di/^cz=d^y 


2ab 


-  to  find  the  values  of  v,  x,  y,  and  z. 


Ans,  - 


x= 


y 


a^d-\-b\-ahd'' 
a'd-hb'c      * 
a^d^-\-a^bc—ab^c 
a'd  +  b^        * 
_  {a^ —ab— hd)hd 
~        aV+6V       • 
_  (cLC -\-bd— a^)acd — (6 — c)b'*c' 
~-  p^  +  6V)rf  * 


SIMULTANEOUS  EQUATIONS,   ETC. 


231 


I*  Given  <  x+z=bi 


to  find  the  values  of  x,  y,  and  z, 

x=\(a-\-h—e). 


7.  Given 


,_y_g=  6, 1 
'—a?— 2=12,  V 
-y-a;=24j 

f    ir4-  y-  2=  8,^ 
<  2a;—  y +32=21,  V   to 
(42+8y-2a;=lY,) 


-4w5.    <  y=i(a— &4-c). 
(2=x(c— «  +  &). 


x—y-'Z=  6, 
6*  Given   •{  3y— a?— 2=12,  }•  to  find  the  values  of  a;,  y,  and  z 

Am.  a;=39,  y=21,  a=12. 


find  the  values  of  ar,  y,  and  z. 
Ans.  x=1,  y=5,  2=4. 


8.  Given  " 


1     l__j5^^ 
a?"^y""6' 
1     1_J^ 

i"^2~T' 

Ly    2    12'J 


••  to  find  the  values  of  x,  y,  and  z. 

Am,  a; =2,  y=3,  2=4. 


f  aj  +  2y  +  32=l'7,  ^ 
9t  Given  <  y  +  22+3a;=13,  >  to  find  the  values  of  ar,  y,  and  z, 
C2  +  2a;  +  3y=12,  ) 

Am.  x=l,  y=2,  2=4. 

ra;+a(y+2)=m,  ) 
10*  Given   <  y  +  b(x-\-z)=nj  >   to  find  the  values  of  ar,  y,  and  2. 
(  2  4-c(a;+y)=^,  ) 

m  +  nca  +pab — na — mcb  —pa 

l  +  2a6c — ba — cb — ca 
n  -{-pah +mbc  —  pb—  nac—mh 

l-\-2ab—ba—cb—ca 

p-\-mbc-\-nca  —  7nc—pba—nc 

l-\-2abc—ba—cb—ca 


Am.  \ 


C      a;—  y+  s=30,  ) 
11.  Given   <    8ar— 4y  +  2s=50,  >   to  find  the  values  of  a?,  y,  and  2. 
C  27a;— 9y4-32=64,  J 

2 
Am.  a?=-,  ^=7,  2=36^-. 
3 


232 


SIMULTANEOUS  EQUATIONS,   ETC. 


'    4x-\-Sr/-\-z     2y-\-2z  —  x-\-\  x—z—5 

— =5+- 


12.  Given  ^ 


10  15  5 

9a;+6y— 225     2x-\- y—Sz_1y-\-z -\-S 
12  4  ~        11 


+i, 


to 


5y  +  30     2a;+3y-2                         3a;  +  2y  +  '7 
— +  J2— y— IH , 


4  12 

find  the  values  of  ar,  y,  and  0. 


6 


^7w.  x=9,  y=*J,  g=3. 


13*  Given  ^ 


>-  to  find  the  values  of  x^  y,  and  z, 

Ans,  a;=120,  y=60,  0=12. 


14,  Given  ^ 


12y— ll2 
11a;— 10y=-^^ , 


>  to  find  the  values  of  ar,  y,  and  2. 


x-\-z  —  2y      z—y—1 
3^       ~        2        ' 
.        3ar       =  y+^  +  V,  J 

^7W?.  «=10,  y=ll,  0=12. 
f  3a;—  y+  0=15,  ') 
15*  Given   <  6a; +  3y— 22=16,  >  to  find  the  values  of  a;,  y,  and  z, 
(  7a;  +  4y— 52=11,  ) 

Ans.  a;=4,  y=2,  2=6. 

r  2a;  +  4y— 32=22,  ) 
16«  Given   <  4a;— 2y  + 52=18,  >  to  find  the  values  of  a;,  y,  and  2. 
(  6a;4-'7y—  2=63,  ) 

Ans.  a;=3,  y=T,  2=4. 

r  3a;  +  2y—  2=20,  ^ 

17.  Given   ^  2a; +  3y  + 62=70,  S  to  find  the  values  of  a;,  y,  and  z. 

I    x—  y  + 62=41,  ) 

Ans.  x=5,  y=6,  2=7. 

:=128,  ) 

18.  Given    ^  3a;+   3y4-72=  60,  >  to  find  the  values  of  a;,  y,  and  2. 

68,  \ 

Ans.  a;=8,  y=5,  2=3. 

f  6a;  +  3y— 42=22,  ) 
19#  Given   <  4a;—  y-f  62=20,  >  to  find  the  values  of  a;,  y,  and  z. 
(5ar  +  2y— 62=11,  ) 

Ans.  a;=3,  y=4,  2=2. 


(  7a;+12y  +  42: 

riven    <  3a;+   3y4-72: 

(ex+     y  +  52: 


SIMULTANEOUS  EQUATIONS,   ETC. 


233 


20.  Given  - 


Sy=u-\-  x+z, 

4z=u-{-  x  +  j/j 

x=u+   14     , 


>■  to  find  the  values  of  u,  x,  y,  and  z. 


Ans.  w=:26,  a;=40,  y=30,  0=24. 


to  find  the  values  of  x,  y,  2,  «, 
and  t 


C        a;—  y—  z=—a,  j 

21.  Given   }  -2x+  y— 22=- a,  >  to  find  the  values  of  a?,  y,  and  0. 

(  — 3a;— 3y+  2=— a,  ) 

Ans,  ar= Jyttj  y=TT^»  z=^ja, 

2x+  ^-22=40,"^ 
4y—  a;  +  30=35, 

22.  Given  ^  3w+    <=13, 

y+  u+   ^=15, 
3a;—  y-\-St—  w=49,^ 

Ans,  x=20,  y=10,  0=5,  w=4,  f=l. 

fM+v  +  a;+y=10,'| 
w  +  i;  +  fl;+s=ll,  I 

23.  Given  -{  u+v+y+z=l2j  }.  to  find  the  values  of  w,v,a;,y,  and  & 
u-\-x-^y+z=lS, 
v+x-\-y+z=l^, 

Ans.  u=l,  v=2,  a;=3,  y=4,  z=6. 


24.  Given 


a;+y 

3 


+  225=      21, 


.        ^±i     -30.= -65, 

^  2 


3aJ4-2/— s         _ 


38, 


<■  to  find  the  values  of  a:,  y,  and  2. 


Ans.  a;=24,  y=9,  2=5. 

^+^y  +  i25=32, 
25.  Given  <  ia;+4-y  +  42=15,  >  to  find  the  values  of  a;,  y,  and  z. 


V  a:+iy  +  i25=32, 
<  Ja;  +  i2/  +  i^=15, 
(  ia;  +  iy  +  i2=12, 


^ws.  x=12,  y=20,  z=SO. 

C        3a:-9y+8.=     41,  )  ^  ^^^  ^^^^  ^^^^^^  ^^ 

26.  Given       -5a:  +  4y  +  2.=  -20, 

f      lla;-7y-62=     37,  ) 

Ans.  x=2,  y=— 3,  2=1. 

(  1x  +  5y  +  2z=^  '79,  ) 

27.  Given  <  8a;+ 7^  +  92=  122,  >  to  find  the  values  of  ar,  y,  and  2. 

(    a;  +  4y  +  62=  56,  ) 

^ws.  a;=4,  y=9,  2=3. 


234 

28.  Given 
Ans,  x= 


SIMULTANEOUS  EQUATIONS,   ETC. 

x-\-y+z=a    ^ 

my=nx  >   to  find  the  values  of  a?,  y,  and  z, 
pz=qx  ) 

anp  amq 


29.  Given 


— 1  y=- 

mp-\-np+mq         mp-\-np+mq 


X     y    a 


,2: 


mp-^np-\-mq 


-  +  -=-  )■  to  find  the  values  of  a?,  y,  and  z. 
X     z     h 


111 

y    z     c\ 


Ans.  X-. 


2abc 


y 


2abc 


z=- 


2abc 


30.  Given 


ac—ab-\-bc'        ab—ac  +  bc'        ac  +  ab—bc 

x+ly-{-  lz=pA 

m>x-\-  y+mz=q^\  to  find  the  values  of  x,  y,  and  z, 
nx+ny+    z=r, ) 


Ans, 


I 


l-l       l-l 


q  m 

1—m      1— m 


-r  1 


m 


1—1 


1  + 


7 


+■ 


m 


■I      1—m 
.       9      , 


1  + 


1— /     1—m      1—n 
I  m  n 


l—n 


I     1- 
9 


l^i 


-  + 


1— » 
r 


1—i 


1  + 


1-^      1— ^ 


1—1 


Note. — This  example  is  the  same  as  Ex.  10,  but  the  answer  is  in  another 
form,  indicating  that  it  has  been  solved  in  a  different  manner.  The  student 
may  observe  that  the  expressions  in  the  brackets  are  identical.  Also,  in  the 
value  of  X,  the  letters  which  are  not  included  in  the  brackets  are  only  those 
which  occur  in  the  first  equation ;  in  the  value  of  y,  only  those  that  occur  in 
the  second  equation ;  and  in  the  value  of  z,  only  those  that  occur  in  the  third 
equation. 

Because  these  values  are  symmetrical,  being  derived  from  symmetrical  equa- 
tions, we  can,  after  getting  the  value  of  x,  deduce  the  value  of  y  from  it,  and, 
having  the  value  of  y,  we  can,  in  hke  manner,  deduce  the  value  of  z.  If  in  the 
ralue  of  a;,  we  change  each  letter  to  that  which  is  in  advance  of  it  in  the  circles, 


QUESTIONS  INVOLVING  SIMULTANEOUS  EQUATIONS.      235 


oa 


we  shall  get  the  value  of  y.  In  like  manner,  changing  the  letters  in  the  value 
of  y,  we  should  get  the  value  of  z.  The  quantities  in  the  brackets  would  stiU 
be  the  same,  though  thus  permuted,  only  the  terms  would  not  occur  in  the 
same  order. 


QUESTIONS    INVOLVING    SIMULTANEOUS 
EQUATIONS    OF    THE    FIRST    DEGREE. 

QUESTION 

(284.)  1.  A  man  and  his  wife  could  drink  a  barrel  of  beer  in  16 
days.  After  drinking  together  6  days,  the  woman  alone  drank  the 
remainder  in  30  days.  In  what  time  would  either,  alone,  drink  a 
barrel? 

SOLUTION. 

Let  a;  =  the  number  of  days  in  which  the  man  could  drink  it, 
And  y  =  «  "  «  woman     "         " 

a;     y     15 
In  6  days  both  drank  j%=^,  leaving  j.     It  took  the  woman  30 
days  to  drink  this  f  ;    therefore,  in  1  day  she  would  drink  ^  of 

Hence,  i=  ^ 
y=  60 

X     50       16 
a;~"l60 


236      QUESTIONS  USrVOLVING  SIMULTANEOUS  EQUATIONS. 

QUESTION 

2.  A  number  consisting  of  2  digits  when  divided  by  4,  gives  a  cer- 
tain quotient  and  a  remainder  of  3  ;  when  divided  by  9,  gives  another 
quotient  and  a  remainder  of  8.  Now,  the  value  of  the  digit  on  the 
left  hand  is  equal  to  the  quotient  which  was  obtained  when  the  num- 
ber was  divided  by  9 ;  and  the  other  digit  is  equal  to  y^  of  the 
quotient  obtained  when  the  number  was  divided  by  4.  What  is  the 
number  ? 

SOLUTION. 

Let  a;  =  the  digit  in  the  tens'  place, 

And  y  =        «        "        units'     " 

Then,  in  consequence  of  our  system  of  notation,  \Ox-\-y  must  rep* 
resent  the  number.  Since  lQx-\-y  divided  by  4  leaves  a  remainder 
of  3,  if  we  subtract  3  from  lOx+y  the  result  is  exactly  divisible  by  4, 
and  by  the  question  the  quotient  is  1*1  y. 

Hence,  we  have  the  equation, 

-^ =17y.    In  like  manner,  we  get 

10a; +  y— 8 

9 =" 

10a;-67y  =3 
10a;+10y  =80 
11y  =11. 
y   =1 
x+1   =8 
X  =1 
Therefore,  71  is  the  number,  because  it  is  equal  tp  lOar+y, 

QUESTION 

3.  What  two  numbers  are  there  in  the  ratio  of  6  to  7,  to  which,  if 
two  other  required  numbers  in  the  ratio  of  3  to  5  be  added,  the  sums 
shall  be  in  the  ratio  of  9  to  13,  and  the  difference  of  these  sums  shall 
be  16  ? 

SOLUTION. 

Let  5x  and  7a;  be  the  numbers  in  the  ratio  of  5  to  7, 
And  3y  and  5y  "  "  ♦♦  3  to  6. 


QUESTIONS  INVOLVING  SIMULTANEOUS  EQUATIONS.       237 

Then,  6ar  +  3y :  Ya;  4- 5y  : :  9  :  13 
Or,   65a;  +  39y=63a;  +  45y 
2a;=6y 

But  (Yaj+Sy)— (6a;  +  3y),  or  2a;  +  2y=16 

a;+  y=8 
Whence,  because  a;=3y,         3y+  y=8 

4y=8 

y=2.      But,  since  x=zZy  and 
y=:2,  we  have  a;=6. 

Hence,  bx  and  7a;,  or  the  first  two  numbers  are  30  and  42  ;  and  3y 
and  6y,  or  the  other  two  numbers  are  6  and  10. 

QUESTIONS. 

1.  What  two  numbers  are  there,  the  greater  of  which  is  to  the  less 
as  their  sum  is  to  42,  and  their  difierence  is  to  6  ?     Ans.  32  and  24. 

2.  A  person  expends  half  a  crown,  or  30  pence,  in  apples  and 
pears,  buying  his  apples  at  4,  and  his  pears  at  5  for  a  penny ;  and 
afterward  accommodates  his  neighbor  with  half  his  apples,  and  one- 
third  of  his  pears  for  13  pence.     How  many  did  he  buy  of  each  ? 

Ans,  72  apples,  and  60  pears. 

3.  A  farmer  sells  to  one  person  9  horses  and  7  cows  for  $600 ;  and 
to  another,  at  the  same  prices,  6  horses  and  13  cows  for  the  same  sum. 
What  was  the  price  of  each  ? 

Ans,  The  price  of  a  cow  was  $24,  and  of  a  horse  $48. 

4.  A  farmer  hires  a  farm  for  $245  per  annum ;  the  arable  land 
being  valued  at  $2  an  Mcre,  and  the  pasture  at  $1.40 ;  now  the  num- 
ber of  acres  of  arable  is  to  half  the  excess  of  the  arable  above  the 
pasture  as  28  :  9.     How  many  acres  were  there  of  each  ? 

Ans.  98  acres  of  arable,  and  35  of  pasture. 
6.  There  is  a  number  consisting  of  two  digits,  the  second  of  which 
is  greater  than  the  first ;  and  if  the  number  be  divided  by  the  sum  of 
its  digits,  the  quotient  is  4 ;  but  if  the  digits  be  inverted,  and  that 
number  divided  by'  a  number  greater  by  2  than  the  difference  of  the 
digits,  the  quotient  becomes  14.     What  is  the  number  1     Ans.  48. 

6.  What  fraction  is  that,  whose  numerator  being  doubled,  and  de- 
nominator increased  by  7,  the  value  becomes  | ;  but  the  denominator 
being  doubled,  and  the  numerator  increased  by  2,  the  value  be- 
comes J  ?  Ans.  f . 


238       QUESTIONS  INVOLVING  SIMULTANEOUS  EQUATIONS. 

Y.  Two  persons,  A  and  B  can  perform  a  piece  of  work  in  16  days. 
They  work  together  4  days,  when  A  being  called  off,  B  is  left  to 
finish  it,  which  he  does  in  36  days  more.  In  what  time  could  each 
do  it  separately  ?  Ans.  ^  in  24  days,  and  -B  in  48  days, 

8.  There  is  a  cistern,  into  which  water  is  admitted  by  three  pipes, 
two  of  which  are  exactly  of  the  same  dimensions.  When  they  are 
all  open,  -^  of  the  cistern  is  filled  in  4  hours ;  and  if  one  of  the 
equal  pipes  be  stopped,  ^  of  the  cistern  is  filled  in  10  hours  and  40 
minutes.     In  how  many  hours  would  each  pipe  fill  the.  cistern  ? 

Ans.  Each  of  the  equal  ones  in  32  hours,  and  the  other  in  24. 

9.  Some  hours  after  a  courier  had  been  sent  from  A  to  B,  which 
are  14*7  miles  distant,  a  second  was  sent,  who  wished  to  overtake  him 
just  as  he  entered  B  ;  in  order  to  do  this  he  must  perform  the  jour- 
ney in  28  hours  less  than  the  first  did.  Now  the  time  in  which  the 
first  travels  IV  miles  added  to  the  time  in  which  the  second  travels  56 
miles  is  13  hours  and  40  minutes.  How  many  miles  does  each  go 
per  hour  ?  Ans.  The  1st  3,  and  the  2d  Y  miles  an  hour. 

10.  ^  and  B  playing  at  backgammon;  A  bet  3  dimes  to  2  dimes 
on  every  game,  and  after  a  certain  number  of  games  found  that  he 
had  lost  17  ^mes.  Now  ha4  A  won  3  more  from  B,  the  number  he 
would  then  have  won  would  have  been  to  the  number  B  would  have 
won  as  6  to  4.     How  many  games  did  they  play  ?  Ans.  9. 

11.  ^  and  B  engaged  to  reap  a  field  of  corn  in  12  days.  The 
times  in  which  they  could  severally  reap  an  acre  are  as  2  :  3.  After ^ 
some  time,  finding  themselves  unable  to  finish  it  in  the  stipulated 
time,  they  called  in  C  to  help  them;  whose  rate  of  working  was 
such,  that  if  he  had  worked  with  them  from  the  beginning,  it  would 
have  been  finished  in  9  days.  Also,  the  times  in  which  he  could 
severally  have  reaped  the  field  with  A  alone,  and  with  B  alone,  are  in 
the  ratio  of  7  to  8.     When  was  C  called  in  ?      Ans.  After  6  days. 

12.  A  vintner  has  2  casks  of  wine,  from  the  greater  of  which  he 
draws  15  gallons,  and  from  the  less  11  ;  and  finds  the  quantities  re- 
maining to  be  in  the  ratio  of  8  to  3.  After  they  become  half  empty, 
he  puts  10  gallons  of  water  into  each,  and  finds  that  the  quantities 
of  liquid  now  in  them  are  as  9  to  5.  How  many  gallons  will  each 
hold  ?  Ans.  The  larger  79,  and  the  smaller  35  gallons. 

13.  At  an  election  for  two  members  of  parliament,  three  men  offer 
themselves  as  candidates,  and  all  the  electors  give  single  votes.     The 


QUESTIONS  INVOLVING  SIMULTANEOUS  EQUATIONS.       239 

number  of  voters  for  the  two  successful  ones  are  in  the  ratio  of  9  to 
8  ;  and  if  the  first  had  had  seven  more,  his  majority  over  the  second 
would  have  been  to  the  majority  of  the  second  over  the  third  as 
12  :  Y.  Now  if  the  first  and  third  had  formed  a  coalition,  and  had 
one  more  voter,  they  would  each  have  succeeded  by  a  majority  of  Y. 
How  many  voted  for  each  ?     Ans.  369,  328,  and  300,  respectively. 

14.  A  wine  merchant  has  two  kinds  of  wine.  If  he  mixes  a  gal- 
lons of  the  first  with  b  gallons  of  the  second,  the  mixture  is  worth  c 
dollars  per  gallon ;  but  if  he  mixes  /  gallons  of  the  first  with  g  gal- 
lons of  the  second,  the  mixture  is  worth  h  dollars  per  gallon.  What 
is  the  price  of  each  kind  of  wine  per  gallon  ? 

(a+b)cg-(f-^g)bh 


The  price  of  the  first  kind  is 
Ans    i  ""^"^^ 

.      u      «      secondkind  (^±^l^-±^ 


dollars 

per 
gallon. 


15.  A  banker  has  two  kinds  of  money ;  it  takes  a  pieces  of  the 
first  to  make  a  crown,  and  b  pieces  of  the  second  to  make  the  same 
amount.  Some  one  gave  him  a  crown  for  c  pieces.  How  many 
pieces  of  each  kind  did  the  person  receive  ? 

Ans.  — — r^  pieces  of  the  1st  kind,  and  -^^ — ~-  of  the  2d  kind. 
a—b    ^  a—b 

16.  What  two  fractions  added  make  |,  and  the  sum  of  whose  nu- 
merators is  equal  to  the  sum  of  their  denominator  ? 

Ans.  I  and  |i. 

17.  A  purse  holds  19  crowns  and  6  guineas.  Now  4  crowns  and  5 
guineas  fill  ^f  of  it.    How  many  will  it  hold  of  each  ? 

Ans.  21  crowns,  or  63  guineas; 

18.  |500  was  to  be  lent  out  at  simple  interest  in  two  separate  sums, 
the  smaller,  at  2  per  cent,  more  than  the  other.  The  interest  of  the 
greater  sum  was  afterward  increased,  and  that  of  the  smaller 
diminished  by  1  per  cent.  By  this,  the  interest  of  tfie  whole  was 
augmented  by  one-fourth  of  the  former  value.  But  if  the  interest  of 
the  greater  sum  had  been  so  increased,  without  any  diminution  of  the 
less,  the  interest  of  the  whole  would  have  been  increased  one-third. 
What  were  the  sums  and  the  rate  per  cent,  of  each  ? 

Ans.  $100  and  $400,  and  4  and  2  per  cent,  respectively. 

19.  Some  smugglers  discovered  a  cave,  which  would  exactly  hold 
the  cargo  of  their  boat ;  viz.  13  bales  of  cotton,  and  33  casks  of  rum. 


240        QtJESTiONS  INVOLVING  SIMULTANEOUS  EQUATIONS. 


1 


"Whilst  they  were  unloading,  a  custom-house  cutter  coming  in  sight, 
they  sailed  away  with  9  casks  and  5  bales,  leaving  the  cave  two- 
thirds  full.     How  many  bales,  or  casks  would  it  hold  ? 

An8.  24  bales,  or  72  casks. 

20.  A  merchant  finds  that  if  he  mixes  sherry  and  brandy  in  quan- 
tities which  are  in  the  ratio  of  2  to  1,  he  can  sell  the  mixture,  at  78 
dimes  a  dozen ;  but  if  the  ratio  be  as  7  to  2,  he  must  sell  it  at  79 
dimes  a  dozen.     What  is  the  price  of  each  per  dozen  ? 

Ans.  Sherry  81,  and  brandy  72  dimes  per  dozen. 

21.  Round  two  wheels,  whose  circumferences  are  as  5  to  3,  two 
ropes  are  wrapped,  whose  difierence  exceeds  the  difference  of  the  cir- 
cumferences by  280  yards.  Now  the  longer  rope  applied  to  the 
larger  wheel  wraps  round  it  a  certain  number  of  times,  greater  by  12, 
than  the  shorter  round  the  smaller  wheel ;  and  if  the  larger  wheel 
turns  round  three  times  as  quick  as  the  other,  the  ropes  will  be  dis- 
chai'ged  at  the  same  time.  What  are  the  lengths  of  the  ropes  and 
the  circumferences  of  the  wheels  ? 

Ans.  The  ropes,  360  and  72  yards ;   and  circumferences  of  the 
wheels  20  and  12  yards. 

22.  If  A  and  B  together  can  perform  a  piece  of  work  in  8  days,  A 
and  C  together  in  9  days,  and  B  and  C  together  in  10  days,  in  how 
many  days  can  each  alone  perform  the  same  work  ? 

Ans.  A  in  14ff  days,  B  in  I7f^  days,  and  (7  in  23/7  days. 

23.  Three  brothers  bought  a  vineyard  for  |100.  The  youngest 
says,  that  he  could  pay  for  it  alone,  if  the  second  would  give  him  1  the 
money  he  had ;  the  second  says,  that  if  the  eldest  would  give  him 
only  the  J  of  his  money,  he  could  pay  for  the  vineyard ;  lastly, 
the  eldest  asks  only  ^  part  of  the  money  of  the  youngest  to  pay  for 
the  vineyard  himself.     How  much  money  had  each  ? 

Ans.  The  oldest  had  |84,  the  2d  |72,  and  the  youngest  |64. 

24.  Three  persons.  A,  B,  and  C,  play  together.  In  the  first  game 
A  loses  to  each  of  the  other  two,  as  much  money  as  each  of  them 
had  when  they  commenced.  In  the  next  game,  B  loses  to  each  of 
the  other  two,  as  much  money  as  they  each  had  at  the  commencement 
of  the  2d  game.  In  the  third  game,  C  loses  to  each  of  the  other  two 
as  much  as  they  each  had  at  the  commencement  of  the  3d  game.  On 
leaving  off,  they  find  that  each  has  an  equal  sum,  namely,  $24.  With 
how  much  money  did  each  commence  ? 

Ans.  A  $39,  B  $21,  and  (7$12» 


Ans. 


QUESTIONS  INVOLVING  SIMULTANEOUS   EQUATIONS.        241 

26.  Three  laborers,  A,  By  and  (7,  are  employed  to  do  a  certain 

piece  of  work.     A  and  B  can  do  the  work  in  a  days  ;  A  and  (7  in  6 

days ;  and  B  and  (7  in  c  days.     How  long  would  it  take  each  to  do 

the  work  alone,  and  how  long  when  they  all  work  together  ? 

2abc        , 

A  requires  -z r  days, 

^  oc+ac — ao 

.  2abc 

B  requires -. days, 

^         bc  +  ab—ac     -^  ' 

.  2abc 

C  requires  — ;-  days, 

^  ab  +  ac—bc      ^  ' 

A.  jB,  and  (7 require  -; r-  days. 

^     '    '  ^        ab-\-ac^bc    -^ 

26.  A  cistern  containing  210  buckets,  may  be  filled  by  2  pipes. 
By  an  experiment,  in  which  the  first  was  open  4,  and  the  second  5 
hours,  90  buckets  of  water  were  obtained.  By  another  experiment, 
when  the  first  was  open  7,  and  the  other  3^  hours,  126  buckets  were 
obtained.  How  many  buckets  does  each  pipe  discharge  in  an  hour? 
And  in  what  time  will  the  cistern  be  filled,  when  the  water  flows  from 
both  pipes  at  once  ? 

Ans.  The  first  pipe  discharges  15,  and  the  second,  6  buckets  in  an 
hour,  and  it  will  require  10  hours  for  them  to  fill  the  cistern. 

27.  A  person  has  two  horses,  and  two  saddles.  The  better  saddle 
cost  $50,  the  other  $2.  If  he  places  the  better  saddle  upon  the  first 
horse,  and  the  worse  upon  the  second,  then  the  latter  is  worth  $8  less 
than  the  other ;  but  if  he  puts  the  worse  saddle  upon  the  first,  and  the 
better  upon  the  second  horse,  the  latter  is  worth  3f  times  as  much 
as  the  former.     What  is  the  value  of  each  horse  ? 

Ans,  The  1st  |30,  and  the  2d  $70. 

28.  A  company  at  an  inn,  expended  a  certain  sum  of  money ;  for 
the  payment  of  which  *they  agree  to  contribute  equally.  Had  there 
been  5  persons  more,  and  had  each  spent  12^  cents  more,  then  the 
bill  would  have  been  $10.17^  ;  but  had  there  been  3  persons  less,  and 
had  each  expended  5  cents  less,  the  bill  would  have  been  $8.25. 
How  many  were  there  in  the  company,  and  what  did  each  spend  ? 

Ans.  There  were  11  persons,  and  each  spent  80  cents. 

29.  A  work  is  to  be  printed  so  that  each  page  may  contain  a  cer- 
tain number  of  lines,  and  each  line  a  certain  number  of  letters.  If 
each  page  should  contain  3  lines  more,  and  each  line  4  letters  more, 
then  there  would  be  224  letters  more  on  each  page ;  but  if  there 

16 


242  INTERPRETATION  OF  NEGATIVE  RESULTS, 

should  be  2  lines  less  on  a  page,  and  3  letters  less  in  each  line,  then 
each  page  would  contain  145  letters  less.  How  many  lines  are  there 
on  each  page,  and  how  naany  letters  in  each  Une  ? 

Ans.  29  lines  in  a  page,  and  32  letters  in  a  line, 

30.  A  coach  set  out  from  Indianapohs  to  Cincinnati  with  a  certain 
number  of  passengers ;  4  more  being  on  the  outside  than  within. 
Seven  outside  passengers  could  travel  at  50  cents  less  expense  than  4 
inside.  The  fare  of  the  whole  amounted  to  $45.  But  at  the  end  of 
half  the  journey,  it  took  up  3  more  outside  and  1  more  inside  passen- 
gers ;  in  consequence  of  which,  the  whole  fare  was  increased  in  the 
ratio  of  17  to  15.  What  was  the  number  of  passengers,  and  the  fare 
of  each  ? 

Ans.  There  were  5  inside,  and  9  outside  passengers.  The  fare  of 
each  inside  passenger  was  $4.50,  and  of  each  outside  passenger 
12.50. 


.^  »«  »  t«  ^  ■ 

INTERPRETATION  OF  NEGATIVE  feESULlS. 

THEOREM. 

(285*)  When  the  value  of  an  unknown  quantity  in  an  equation 
is  found  to  be  negative,  this  result  indicates  that  an  absurdity  is  in- 
yolved  in  the  enunciation  of  the  problem. 

DEMONSTRATION. 

Find  a  number  which,  added  to  +  4,  makes  2, 

Let  X  =  the  required  number.  * 

Thenar-|-4=2 

a;=-2 

This  result  indicates  that  the  problem  was  incorrectly  enunciated ; 
the  words  added  to  being  used  for  subtracted  from.  The  problem  is, 
however,  worded  correctly,  if  we  consider  the  word  added  in  its  ex- 
tended algebraic  sense.     Let  us  word  this  problem  differently : 

Find  a  number  to  which  if  4  be  added  the  sum  will  be  2. 

The  result  x=—2  indicates  that  the  problem  is  absurd  in  an 
arithmetical  sense,  but  definite  in  an  algebraic  sense. 


INTERPRETATION  OF  NEGATIVE  RESULTS.  248 

Let  us  take  another  problem : 

A  father  whose  age  is  42  years  has  a  son  whose  age  is  12.     In  how 

many  years  will  the  aye  of  the  son  be  I  that  of  the  father^ 

This  question  gives  the  equation 

,^     42+aJ 
a;4-12=— — -. 
4 

The  solution  of  which  gives  «=— 2. 

This  result  shows  that  the  son's  age  will  never  be,  in  the  future^ 
J  the  father's  age ;  but  that  2  years  ago  the  son's  age  was  \  that  of 
the  father. 

That  part  of  the  question  which  we  have  put  in  italics  should  have 
been,  How  many  years  since  the  age  of  the  son  was  I  that  of  the 
father^ 

Let  us  take  still  another  example. 

A  man  worked  1  days,  and  had  his  son  with  him  3  days ;  and 
received  for  wages  $2.20.  He  afterward  worked  5  days,  and  had  his 
son  with  him  1  day,  and  received  for  wages  $1.80.  What  were  the 
father's  daily  wages,  and  what  was  the  effect  of  the  son's  presence  ? 

Letting  x  =  the  father's  daily  eflfect, 

And      y  =    "   son's        " 

We  get  the  equations,      72:4-3^=220,  )  , 

and  5X+  y=180,  S  ^  ^' 
provided  the  daily  effect  of  each  is  productive  of  wages. 

But,  if  the  daily  effect  of  the  father  adds  to  the  amount  of  wages, 
and  the  daily  effect  of  the  son  diminishes  the  amount  of  wages,  the 
equations  must  be 


and 


1x-Sy=220,]     . 
5x—  y=180.  P  ''' 


If  the  reverse  were  true,  we  should  have 
-1x+Sy=220,) 
and  —5x+  y=l80.)^  ^' 

Which  of  these  three  couplets  is  the  one  which  belongs  to  this 
problem  ?  This  question  may  be  answered  after  we  have  found  the 
values  of  x  and  y  in^ach. 

The  (1)  gives  a;=40  and  y=— 20;  the  (2),  a;=40  and  y=20; 
and  the  (3),  ir=— 40,  y=  —  20. 

Now,  the  (3)  results  can  not  be  true,  because  they  make  the  daily 
effect  of  each  dicffinish  the  amount  of  wages,  while  the  equations  from 
which  these  values  were  derived  considered  this  to  be  true  only  of 
the  father's  daily  effect 


244  INTERPRETATION  OP  NEGATIVE  RESULTS. 

The  (1)  results,  in  like  manner,  indicate  that  the  daily  effect  of 
the  son  is  to  diminish  the  amount  of  wages,  while  the  equations  from 
which  they  were  derived  considered  both  to  be  productive  of  wages. 

The  results  derived  from  the  (2)  couplet  are  the  only  ones  that 
fulfill  the  arithmetical  conditions.  The  form  of  the  (2)  couplet,  also, 
shows  how  the  question  should  be  stated ;  or,  in  other  words,  that 
the  son's  presence  diminished,  each  day,  the  father's  earnings  by 
an  amount  equal  to  20  cents. 

We  may  suppose  the  father  had  to  allow  his  employer  20  cents 
a  day  for  his  son's  board ;  or  that  he  was  hindered  one-half  a  day  in 
his  work  every  day  the  son  was  present ;  or  that  the  father,  out  of  his 
own  wages,  requested  the  employer  to  allow  the  son  20  cents  for 
every  day  the  son  worked. 

Guided  by  such  ideas,  we  are  enabled  generally  to  give  an  arith- 
metical explanation  of  negative  results. 

QUESTIONS. 

1.  A  man,  at  the  time  of  his  marriage,  was  60  years  old,  and  his 
wife  40.     When  was  he  twice  as  old  as  she  ? 

Ans.  30  years  before  marriage. 

2.  What  fraction  is  that  which  becomes  |  when  1  is  added  to  its 
numerator,  and  becomes  ^  when  1  is  added  to  its  denominator  ? 

Ans.  5^. 

3.  A  man,  when  he  was  married,  was  30  years  old,  and  his  wife  15. 
How  many  years  must  elapse  before  his  age  will  be  three  times  his 
wife's  age. 

Ans.  He  was  three  times  as  old  as  she  Y^  years  before  their  marriage. 

4.  What  fraction  is  that  which,  if  2  be  added  to  its  numerator,  its 
value  is  zero ;  but,  if  6  be  added  to  its  denominator,  its  value  is  infinite  ? 

Ans.  — -. 
—  6 

6.  What  fraction  is  that  which,  if  3  be  added  to  its  numerator,  its 
value  is  nothing;  but,  if  4  be  subtracted  froni  its  denominator,  its 
value  is  1  ? 

Ans.  — -. 
%  1 

6.  A  laborer  working  for  a  gentleman  during  12  days,  having  with 
him,  the  first  7  days,  his  wife  and  son,  who  occasion  an  expense  to 


GENERAL  DISCUSSION,    ETC. 


245 


i 


him,  received  $4.60 ;  he  afterward  worked  8  days,  during  5  of  which 
his  wife  and  son  were  with  him,  and  received  $3.00.  What  were  the 
wages  of  the  laborer  per  day,  and  also  the  expense,  per  day,  of  his 
wife  and  son  ? 

Ans,  His  daily  wages  50  cents,  and  expense  of  wife  and  son  20 
cents. 

7.  Two  men,  A  and  £,  commenced  trade  at  the  same  time ;  A  had 

3  times  as  much  money  as  B ;  A  gained  $400,  and  ^  $150 ;  now, 
A  has  twice  as  much  money  as  £,    How  much  had  each  at  first  ? 

Ans,  A  was  in  debt  $300,  and  £  $100. 

8.  A  man  worked  10  days,  his  wife  4  days,  and  his  son  3  days, 
and  their  wages  amounted  to  $11.50  ;  at  another  time,  he  worked  9 
days,  his  wife  8  days,  and  his  son  6  days,  and  their  wages  amounted  to 
$12.00 ;  a  third  time,  he  worked  7  days,  his  wife  6  days,  and  his  son 

4  days,  and  their  wages  amounted  to  $9.00.  What  were  the  daily 
wages  of  each  ? 

Ans.  Husband's  daily  wages,  $1.00 ;  wife's,  0  ;  and  son's,  50  cents. 

9.  The  sum  of  two  numbers  is  120,  and  their  difference  is  160 ; 
what  are  the  numbers  ?  Arts,  140  and  —20. 

10.  What  number  is  that  whose  |  part  exceeds  its  i  part  by  12  ? 

Ans.  —144. 


i^  «>  ♦  >.  »■ 


GENERAL  DISCUSSION  OF  CERTAIN  RESULTS  OBTAINED 
IN  THE  SOLUTION  OF  SIMPLE  EQUATIONS. 

PROBLEM. 
(286.)  To  interpret  the  result  0=0. 


SOLUTION 
Let  us  endeavor  to  solve  the  problem  : 


Given 


0=0 


>  to  find  the  values  of  x  and  y. 


(3)=(l)xa 
(4)^(2)-(3) 


246  GENERAL  DISCUSSION,   ETC. 

Hence,  We  ate  unable  to  obtain  the  values  of  x  and  y. 
By  looking  at  the  second  equation,  we  see  that  it  is  not  independent 
of  the  first,  it  being  the  first  multiplied  by  a. 

We  have,  then,  only  ^H — =c  to  find  the  values  of  x  and  y. 

We  have  already  shown  that  when  there  are  two  unknown  quan- 
tities and  but  one  equation,  the  values  of  x  and  y  are  indeterminate, 
or,  in  other  words,  that  there  are  an  infinite  number  of  corresponding 
values  of  x  and  y,  which  will  satisfy  the  equation. 

Again,  let  2a?  +  a= a;  4- 1« + ^  +  !«. 

This  equation,  also,  reduces  to  0=0.  This  ought  to  be  the  case, 
because  2x  +  a=2x  +  a,  or  2a;=2a;,  or  a:=a;,  is  an  indeterminate  equa- 
tion, since  x  may  be  any  quantity  whatever. 

Hence,  the  result  0=0  is  a  sign  of  indetermination,  and  when  obtained 
from  two  simultaneous  equations,  indicates  that  there  is  but  one  con- 
dition.   When  the  values  of  x  and  y  are  indeterminate,  it  is  customary 

to  indicate  it  by  ^=x  and  y=r« 

PRO  BLEM. 

(287.)  Find  a  number  such,  that  if  9  be  added  to  8  times  the 
number,  and  this  sum  be  divided  by  6,  the  quotient  will  be  equal  to 
13  augmented  by  12  times  the  number,  and  this  sum  divided  by  9. 


SOLUTION. 

Let  X  =  the  required  number. 

Then  we  have,  by  the  conditions  given, 

8a;+9     13  + 12a: 
6               9 

(1) 

'?2a;  +  81  =78   -}-72a; 

(2)=(l)x64. 

'72a;— 72a:=78   —81 

81   -78  =l2x-'I2x 

3   =0 
This  result  is  an  absurdity,  and  indicates  that  the  conditions  of  the 
question  are  incompatible  or  contradictory. 
The  conditions,  we  have  seen,  give 

72a: +  81  =  72a; +  78. 
This  may  assume  the  form 

72a;  +  78  +  3  =  72a;  +  78. 
This  equation  can  be  satisfied  only  on  the  condition  3=0;  but  3 
does  not  equal  0,  and  therefore  the  equation  is  impossible. 


GENEEAL  DISCUSSION,    ETC.  ^7 


PBOBLEM. 
(288.)  To  interpret  x=^ . 

SOL  UTION. 
We  may  consider  that  oo==-  when  a=iO,  and,  also,  00 =-7  when 
1 
b=0.    Then,  ^=— ,  becomes  iP=5Q-,  when  both  a  and  h  are  boA- 


b 
equal  to  0. 

a    1     I     I     h    h 
But  {»=-=— f--=-x-=-. 
1     a     b     a     I     a 

h 

Now,  if  we  make  a  and  b  each  equal  to  0,  we  have  x=7> 

0 
We  see  by  this  that  we  can  get  ^=S  and  «=-  from  the  same 

1 

equation,  a;=-,  by  making  both  a  and  b  equal  0 ; 

Hence,  ^  is  equivalent  to  -,  or,  in  other  words,  is  sometimes  a 

symbol  of  indetermination,  for  we  have  already  shown  that  -  is 
sometimes  a  sign  of  indetermination.  In  vanishing  fractions  the  value 
of  -  is  determinate. 

PROBLEM. 
(289.)  To  interpret  x=0  .  00  . 

SOLUTION. 
Letting  ar=ax -7,  and  making  both  a  and  b  equal  to  0,  we   get 

a;=Ox-,  or  a;=:0  .  00. 

But  x=a  X  T  is  the  same  as  a:=Tj  or,  when  a  and  b  are  both  equal 

to  0,  x=-. 


248  GENERAL   DISCUSSION,   ETC. 

We  see,  then,  that  0  .  oo  is  equivalent  to  -. 

PROBLEM. 
(290.)  To  interpret  a;=  oo  —  go. 

SOLUTION. 

Let  (c= -. 

a     0 

If,  in  this  equation,  both  a  and  h  are  0,  we  have 

1     1 
x=-—~,  or  x=  00  —  00 . 

But  a:= 7  is  the  same  as 

a     0 

h — a 

x=z . 

ab 

Which  equation  becomes,  when  a  and  h  are  each  equal  to  0, 

_0-0 

0 

or,  ^=3. 

0 

We  see,  then,  that  00  —  00  is  equivalent  to  -. 

PROBLEM. 
(291.)  Two  couriers  traveled  on  the  same  road;  when  one  was 
at  A  the  other  was  at  B,    When  were  they  together  ? 

SOLUTION. 

Let  '  ^  represent  the 

road  and  the  given  places.  This  problem  is  stated  in  very  general 
language,  and,  therefore,  embodies  many  conditions.  It  will,  however 
be  found  not  to  be  so  general  as  the  algebraic  formula  to  which  it 
gives  rise. 

To  obtain  this  formula  let  us  take  one  of  the  many  cases  which 
may  occur,  viz.,  that  the  faster  traveler  was  at  A  when  the  other 
was  at  Bj  and  that  they  were  together  at  D. 

A  B  D 

Let  a  =  the  distance  from  Ato  B  ;  m  •=.  the  number  of  miles  the 
faster  traveler  went  per  hour ;  and  n  =  the  number  of  miles  the  other 
went  per  hour. 


GENEBAL  DISCUSSION,   ETC.  249 

Let  x=^AB^  and  y—BD. 

We  have  then  —  the  number  of  hours  that  it  took  the  faster  cour- 
m 

ier  to  ffo  the  distance  x,  and  -  =  the  number  of  hours  it  took  the 

^  n 

other  to  go  the  distance  y.     Since  these  terms  must  be  equal,  we  have 

X y 

m  ~n 

But  x—y^a 

The  solution  of  these  equations  give 

am 
x=. 


and  y= 


m—'i 
an 


m—'i 


X  XI 

Whence,  — ,  or  ■-: 


m        n     m—n 

These  results  are  applicable  to  a  great  number  of  separate  supposi- 
tions, or  we  may  consider  them  as  the  solution  of  the  general 
problem. 

In  order  to  adapt  these  results  to  all  the  supposable  cases,  we  must 
first  establish  certain  conventional  signs. 

When  m  is  a  positive  quantity,  we  shall  consider  that  the  traveler 
who  went  m  miles  per  hour,  went  eastward,  or  to  the  right ;  but  when 
w  is  a  negative  quantity,  we  shall  consider  that  he  went  westward,  or 
to  the  left.  We  shall,  also,  attach  the  same  ideas  to  n.  When  x  is 
a  positive  quantity,  we  shall  consider  that  D  is  the  eastward,  or  to  the 
right  of  A ;  but  when  it  is  a  negative  quantity,  we  shall  consider  that 
D  is  westward,  or  to  the  left  of  A.  Also,  we  shall  consider  D 
as  eastward  or  westward  from  B,  according  as  y  is  positive  or 
negative. 

X  11 

When  — T  or  -  is  positive,  we  shall   consider  the  travelers   were 

X  11 

together  after  they  were  at  A  and  J?,  but  when  —  or  -  is  negative, 

that  they  were  together  before  they  were  at  A  and  B.  But  how  shall 
we  establish  the  meaning  of  4- a  and  — a,  since,  according  to  the 
above  ideas,  the  distance  a  would  have  a  different  sign  according  as 
^e  refer  it  to  ^  or  i?  ? 

We  shall  let  the  point  A  rule ;  and  shall  call  AB^  or  a,  positive 


250 

when  B  is  to  the  right  of  A  ;  and  AB,  or  a,  negative  when  5  is  to 
the  left  of  A,    Thus 

-k i 

'  ■ 

B  A 

"We  shall  consider  AB  in  the  first  equal  to  +a,  but  in  the  second 
line  equal  to  —a. 

"With  conventional  ideas,  we  can  make  a  particular  problem  out  of 
each  of  the  following  suppositions,  and  the  corresponding  values  de- 
duced from  the  above  formulas  will  be  the  correct  results  for  each 
particular  supposition. 

1st  When  m=+2,  w=-fl,  and  a=+3,  we  get  a;=+^,  and 

y  =  4-  3,  — =  +  3,  and  -=  +  3. 

2d.  "When  m=:— 2,  w=  — l,and  a=+3,  we  get  a;=  +  6,  y=+3, 

X  V 

—=—3,  and  -=  —  3. 
m  n 

3d.  When  m=4-2, 7i=  — 1,  and  a=+8,  we  get  x=:  +  2,  y=  — 1, 

— =+1,  and  -=  +  1. 
m  n 

4th.  "When  m=—2,  ?i=  +  l,  and  a=-|-3,  wegeta;= -}-2,y=  — 1, 

— =-l,and  ^=-1. 
m  n 

5th.  "When  m=+2, 7i=  +  l,and  a=— 3,we  get  ar=— 6,  y=— 3, 

X  V 

—=—3,  and  -=—3. 
m  n 

6th.  "When  m=— 2,w=— l,and  a=— 3,  we  get  aj=— 6,  y=--3, 

X  V 

— =+3,  and -=+3. 
m  n 

Yth.  "When  wi=  4-2, 71=  — 1, and  a=:— 3,we  get  a;=— 2,  y=  +  l, 

-^=-1,  and^=-l. 
m  n 

8th.  When  m=  — 2,  w=  +  l,and  a= — 3,  we  get  a;=  — 2,  y=  +  l, 

X  V 

— =  +  1,  and  :^=:  +  l. 
m  n 

In  the  first  four  suppositions  B  is  east  of  A^  and  in  the  last  four,  B 

is  west  of  A. 

■  '       '  ' 

A        D'    B  D 

In  the  first  two  suppositions  the  couriers  were  together  at  Z>,  in  the 


GENERAL  DISCUSSION,    ETC.  251 

Ist  3  hours  after  they  were  at  A  and  B^  and  in  the  2d  3  "before.  In  the 

next  two  suppositions  they  were  together  at  D\  in  the  3d  1  hour  after 

they  were  at  A  and  B^  and  in  the  4th  1  hour  before, 

'  '       ' 

D"  B    D'"        A 

In  the  next  two  suppositions,  they  were  together  at  D\  in  the  5th 
3  hours  before  they  were  at  A  and  -6,  and  in  the  6th  3  hours  after. 
In  the  next  two  suppositions  they  were  together  at  D"\  in  the  Vth  1 
hour  before  they  were  at  A  and  B^  and  in  the  8  th,  1  hour  after. 

9th.  When    m=-|-c,    w=+c,    and    a=4-c?,   we  get  ic=:-fQO, 

y=  -f-  oo,  — =  -}-QO  ,   and  -=  +  Qo. 
m  n 

10th.  When   m=—c^  n=—c,  and    a=z+d^  we   get    x=+  oo, 

y=+oo  ,  — =—  oo,  and -=—00  . 
m  n 

11th.  When  m=:  +c,  n=  — c,  and  a=  +c?,  we  get  x=  +-,  y=  — -, 

a;  c?        .  y       '  d 

——  +— ,  and  -=  +— . 
m         2c  n         2c 

12th.  When  m=  — c,  w=  +  c,  and  a=  +  a,  we  get  a:=  +  -,  y  =  — -, 

2  2 

X  __^      d  y  _      d 

m~     2c*         n"     2c' 

13th.  WTien   m=+c,  w=+c,    and   a=—dy  we   get  a;=  — cx), 

y=— oo, — =— QO  ,  and  -=— oo. 

14th.  When   m=:—c,  n:=z-^c,  and   a=—dj  we   get  «=—  oo, 

X  1/ 

y=— oo, — =+oo  ,  and  -=  4- oo. 

15th.  When  m=:  +  c,  w=  — c,  and  a=  — tf,  we  get  x=  — -,  y  =  +  -, 

i2  2 

a?  _      (?  y_      c? 

w"     2c'  w         2c* 

rf  d 

16th.  When  m=— c,  w=  +c,  and  a=  —a,  we  get  x=  — -, y=  +-, 

Z  2 

X  d         .  y  d 

— =+— ,  and-=+— . 
m         2c  n         2c 

^  D""       B 

In  the  9th  suppo^+ion  the  couriers  will  be  together  at  a  point  jn- 


252  GENERAL   DISCUSSION,    ETC. 

finitely  distant  to  the  east  of  A  and  B  in  an  infinite  number  of  hours 
from  the  time  they  were  at  A  and  B.  In  the  10th  supposition  they 
were  together  at  the  same  place  an  infinite  number  of  hours  before 
they  were  at  A  and  B,  The  reverse  of  these  results  is  found  in  sup- 
positions 13  and  14.  To  say  that  two  couriers  were  together  an  in- 
finite number  of  hours  ago,  is  the  same  as  saying  they  never  were 
together,  and  to  say  they  will  be  together  in  an  infinite  number  of 
hours,  is  the  same  as  saying  they  never  will  be  together. 

It  should  be  remembered  that  in  the  13th  and  14th  that  A  is  east 
oiB, 

It  will  be  seen  that  to  obtain  these  infinite  results,  we  have  consid- 
ered 0  at  one  time  as  +0,  and  at  another  —0.  We  started  out  with 
the  idea  that  the  position  A^  or  that  of  the  faster  traveler,  should  rule 
the  sign  of  a.  But  wfien  both  travel  equally  fast,  there  seems  to  be  a 
difficulty.  To  remove  this,  we  consider  m  disregarding  its  sign- as 
greater  than  n  by  an  infinitely  small  quantity.  Hence,  7?i— w=+0 
and  — m  +  w= — 0. 

In  suppositions  11,  12,  15,  and  16,  the  couriers  were  together  at 
i)"",  a  point  midway  between  A  and  j5,  A  being  the  east  point  in  11 
and  12,  and  the  west  point  in  14  and  15.     In  these  four  suppositions, 

the  number  of  hours  was  the  same,  —  hours.  +  —  indicating  that 
they  were  together  —  hours  after  they  were  at  A  and  B^  and  —  —  that 
they  were  together  —  hours  before  they  were  at  A  and  B. 

lYth.  When  m=+c,  w=+0,  anda=+Q^,we  geta;=+(f,  y=-f  0, 

X          d       ^  y         0 

— =+-,  and^=+-. 
m         c  n         0 

18th.  When  m=— c,7i=— 0, and  a=. +(^,wegeta;=4-<?, y=4-0^ 

X         d       ^  y         0 

— = ,  and  -=  — -. 

m  c  n         0 

19th.  Whenm=+c,  %=—0,  and a=+cf,  we  get  a:= -he?,  y=—0, 

X  d       .y        0 

— =+-,  and -=-!--. 
m      ^c'         n         0 

20th.  When  m=  — c,  w=  -1-0,  and  a=  -he?,  we  get  «=  -|-c?,  y—  —0, 
^  ^     d         y  __     0 


253 

21st.  When  m=  +c, /i=+0,  and  a=.—d,  wegetx=—djy=-'0, 

m~     c'         n~     0* 

22d.  When  m=  — c,  w=:  — 0,  and  a=  —d,  we  get  a;=  — c?,  y=  —0, 

X         d       ^y        0 
— =+-,  and -=+-. 

23d./Wlien  m=  +c,  w=  —0,  and  a=  — <?,  we  get  y=  — <?,  y=  +0, 

^ _/ ^    ^y_   ^ 

m  ~/     c'         ?i~     0* 
^24th.  When  m=  — c,  7i=  +  0,  and  a= — c?,  we  get  a?=  — c?,  y=  +  0, 

^  d       ^y         0 

— =+-,  and  -=+-. 

y         0  y         0  y 

In  these  results  we  get  -=4--,  and  -=— -.    But -is  not  there- 
*=     w         0  w         0  n 

fore  indetenninate,  because  -= — ,  and —  is  either  +-  or  — . 

n     m  m  c  c 

»=  4-0  indicates  that  the  courier  who  was  at  B  stood  still  with  his 
face  to  the  east,  and  n=—0  that  he  stood  still  with  his  face  to  the 

west. 

25th.  When  m=  +0,  n=  4-0,  anda=  +d,  we  geta;=  4--,  y=z  4--, 

0  0 

X         0       ,  y     0 
— =4--,  and  -=-. 

26th  Whenm=— 0,  n=— 0,  and  a=  f  c?,  we  get  a;=4-r,  y=+T, 


«         0       ,  y        0 
m  0'         /i         0 

2*7 th.  When  m—  4-0,  n=  —  0,  and  a=  +d,  we  get  ar=  4--,  y=  — -, 

.0  0 

a:  0        ,  y         0 

—=+-,  and^=4--. 
m  0  n         0 

28th.  When  m=: — 0,  w=  4-  0,  and  a=  + 1?,  we  get  x=  4-  j:^  y  =  ~ z, 
0  0 

^— —?      ,y__o 


264  ^  GENERAL  DISCUSSION,   ETC. 

29th.  When  m=  +  0,  n=  -f  0,  and  a=  —d,  we  get  x=  — ,  y  =  — , 

0  0 

±=_2  and^=-- 
m         0*         w~     0* 

SOth.  When  m=  —0,  »=  —0, and  a=  — <^,  we  get  x=  — -,  y=  — - 

0  0 

a;  0       ,  y         0 

—=+-,  and-=+-. 
m         0'        w        0 

31st.  When  m=  +0,  w= —0,  and  a=  — rf,  we  get  ir=  — ,  y=  H — , 


^=4  and  ??=-?. 
m  0  «         0 

32d.  Whenm=— 0, 7^=4-0,  anda=:--(f,  we  get  a:=---,  y=  + 

0  Kj 

0  0 

a;         0       ,  y        0 

—=  +  -,  and-=+-. 

These  results  are  peculiar,  and  are  not  indeterminate  in  the  sense 
usually  attached  to  the  word  indeterminate,  since  its  technical  mean- 
ing is  that  there  is  an  infinite  number  of  values  which  will  satisfy  the 
given  conditions.  But  there  is  an  absurdity  in  asking  when  they 
were  together  if  there  was  a  distance  between  them  and  each  stood 
still.  We  should  obtain  the  same  results  if  at  the  same  time  a  was 
made  +0or  —  0,  and  then  they  would  be  the  true  symbols  of  indeter- 
mination. 

I^  in  the  first  eight  suppositions,  we  put  +0  instead  of  +  3,  and  —0 
instead   of   —3,  we  get  ic=:-l-0   or  0?=— 0,  y=+0  or  y=— 0, 

X  OS 

— =+0  or  — -  =  — 0,  which  results  are  easily  interpreted. 

If,  in  the  second  eight  suppositions,  we  put  a=4-0  or  a=r— 0,  we 

0  0  0  0,a;0a?0 

geta:=+-or^=--,y=+-,ory=--,and--+-,or-=— , 

which  are  true  symbols  of  indetermination,  showing  that  they  were 
always  together  and  always  will  be. 

If,  in  the  third  eight  suppositions,  we  put  a=  4-0  or  a=— 0,  we 


GENEEAL  DISCUSSION,   ETC. 


255 


get  aj=+0  ora;=:— 0,  y=+0  or  y=— 0,  — =+0  or  — =— 0,  and 


m 


m 


y        0      y        0 

-=  +-  or  -=— -,  which  results  are  easily  interpreted. 
n         0       n  *     0 

We  have  then  made  64  suppositions,  all  of  which  admit  of  a  simple 
interpretation  except  eight,  which  embodied  an  absurdity. 

These  suppositions  and  results  show  that  algebraic  formulas  are  far 
more  extensive  than  the  particular  problem  from  which  they  may  be 
derived. 


EXAMPLE  S 


1,  Given 


(  6x-\-   9^=27,  ) 
(  8a;  +  12y=36,  i" 


to  find  the  values  of  x  and  y. 

A  0         0 


2«  Given  j  —qI  ^  ^^^  ^®  values  of  a:  and  y. 


A  0         0 

Ans,  x=-.  y=-. 

0'^     0 


3«  Given 


-  to  find  the  values  of  x  and  y. 


f  2x-  y=l,  ^ 

5x—2i/=4y 

3x-\-   y=9, 

L  Sx-  y=2,  J 

Ans.  No  value  can  be  found  that  will  at  the  same  time  satisfy  all 
the  equations. 

"  2a;-  y=:l,  " 
5ar— 2y=4, 
3x+  y=9, 
3a;-  y-S, 

Ans.  a:=2,  y=3.    Explain  these  results. 
(    x+  y+   2=3,^ 
5.  Given   <    x—  y—'z=lA  to  find  the  values  of  a?  and  y. 
/  2a;4-2y  +  22=l.  \ 


4«  Given  ^ 


-  to  find  the  values  of  x  and  y. 


Ans.  x=-,  y=—  Qo,  z=+  oo. 
Ca;  +  y  +  22=     2,^ 
6.  Given   <x-\-y-\-2z=     1,  >  to  find  the  values  of  ar,  y,  and  0. 
(a;+y  +  22=-l,  ) 

A  0         0         0 

0 


256  SIMPLE   mEQUATIONS. 


Note. — ^The  results  in  the  last  two  examples  were  obtained  by  tbe 
checker-board  process,  and  show  that  -  is  not  always  a  symbol  of  ir 
determination,  for  in  these  examples  the  equations  ar^  incompatible. 


•  ♦>'»■ 


SIMPLE   INEQUATIONS. 

(29 2«)  An  iiiequation  is  that  which  denotes  that  one  algebraic  ex- 
pression is  greater  or  less  than  another.  Thus  a>6,  and  c<C^d  are 
inequations,  which  denote  that  a  is  greater  than  6,  and  that  c  is  less 
than  d. 

(293.)  Two  inequations  subsist  in  the  same  sense  when  the 
sign  of  inequality  has  the  same  position  in  both.  Thus  a>6  and 
c>c?  are  inequations  in  the  same  sense  ;  so  also,  are  a<6  and  c<(?. 

(294.)  Two  inequations  subsist  in  a  contrary  sense  when  the  sign 
of  inequality  has  not  the  same  position  in  both.  Thus  o>6  and  c<(/ 
are  inequations  in  a  contrary  sense ;  so  also,  are  a<6  and  c>c?. 

THEOREM. 

(295.)  The  addition  or  subtraction  of  an  equation  to  or  from  an 
inequation  results  in  an  inequation  in  the  same  sense. 

THEOREM. 

(296.)  The  addition  of  two  inequations  which  subsist  in  the 
same  sense,  results  in  an  equation  in  the  same  sense. 

THEOREM. 

(297.)  The  subtraction  of  an  inequation  from  another  which 
exists  in  the  same  sense,  does  not  necessaiily  result  in  an  inequation  in 
the  same  sense. 

DEMONSTRATION. 

Subtracting  4>3  from  8>5,  we  get  4> 2  which  is  true. 

But  if  we  subtract  8>5  from  4>3,  we  get  —  4>— 2,  which  is 
not  true  according  to  the  conventional  idea  sometimes  attached  to 
negative  quantities,  that  they  are  less  than  nothing. 


SIMPLE  INEQUATIONS.  257 

But  this  inequation  is  still  true  if  we  limit  the  reasoning  to  negative 
quantities,  for  —4  is  evidently  a  greater  negative  quantity  than 
—  2.     A  debt  of  4  dollars  is  greater  than  a  debt  of  2  dollars. 

On  the  other  hand,  we  say  that  a  man  who  is  in  debt  2  dollars,  and 
has  nothing,  is  worth  more  than  the  man  who  is  in  debt  4  dollars  and 
has  nothing. 

Hence  considered  in  the  light  of  wealth  —2  is  greater  than  —4, 
or  in  other  words,  —2  comes  nearer  being  a  positive  quantity  than 
—4  does.  Again,  let  us  subtract  8>2  from  11  > 9,  and  we  have 
3  <  7,  an  inequation  which  subsists  in  a  contrary  sense.  Let  us  put 
these  inequations  in  a  different  form : 

9  +  2>9 
and     2  +  6>2 

Subtracting  we  get  —  4>0,  dropping  equal  quantities  from  both 
sides.  But  we  have  just  seen  that  this  inequation  must  subsist  in  a 
contrary  sense,  that  is  —  4<0.  This  shows  —4  must  be  considered 
less  than  nothing. 

We  can  also  prove  this  as  follows : 

Since  3<7       it  follows  that 

3-'7<'7-'7 
-4<0 

THEOREM. 
(298.)  The  multiplication  or  division  of  an  inequation  by  a  posi- 
tive quantity,  results  in  an  inequation  in  the  same  sense. 

THEOREM. 
(299»)  The  multiplication  or  division  of  an  inequation  by  a  nega- 
tive quantity,  results  in  an  inequation  in  a  contrary  sense. 

DEMONSTRATION. 
Let  a>&.     Now  if  we  multiply  this  by  —  c,  we  get  —ac<^—bc, 
because  the  greater  a  negative  quantity  is  numerically  the  less  it  is 
considered. 

Thus,  4<5  gives,  by  multiplying  by  —6,  — 24>  — 30. 
Since,  multiplying  by   —1  is  equivalent  to  changing  signs,  we 
derive  the 

COROLLARY. 

WTien  the  signs  in  both  members  of  an  inequation  are  changed^  the 
sign  of  inequality  must  be  reversed. 

IT 


258 


SIMPLE   INEQUATIONS. 


Thus  clianging  the  signs  in  —a  +  b—c'^d—m,  we  get 

a  —  b  +  c<^m—d. 


THEOEEM. 


(300.)  When  both  members  of  an  inequation  are  positive,  they 
may  be  raised  to  the  same  power,  and  the  result  will  be  an  inequation 
existing  in  the  same  sense.  ^ 


THEOREM. 


(301*)  When  both  members  of  an  inequation  are  positive,  the 
same  root  of  each  may  be  extracted  and  the  result  will  be  an  inequa- 
tion existing  in  the  same  sense. 


PBOBLE 


(302*)  Find  the  limit  to  the  value  of  x  in  the  inequation 
V.-f  >|+5  (1). 


SOLUTION 


21a;— 23>2a;+15 
19a;>38 
iP>  2 


(2)=(l)x3 

(3) =(2)  transposed, 

(4)=(3)-19. 


EXAMPLES. 

!•  Given  x  +  ^x-\-^x^ll  to  find  the  limit  of  x.         Ans.  ic>6. 
2.  Given  S^c  +  Yar— 30>10  to  find  the  limit  of  x.       Ans,  ar>4. 


I.  Given 


> 


to  find  the  limit  of  x,         Ans.  a;>2c?. 


ed    ^      Sx 

4«  Given  ^a;  +  3a;— 5>16  to  find  the  limit  of  x.         Ans.  a;>6. 
5«  Given  Ya;— 1>34  to  find  the  limit  of  x.  Ans.  a:>5. 

6*  Given  i  ^       .  ^  . «     ^'    r  to  find  an  intej2:er  value  of  x. 
(  2a;  +  4>16  — 2a;,  )  * 

Ans.  ar=4. 

••  to  find  the  limit  of  x. 

Ans.  a;>a  and  a;<6. 


7.  Given 


ax     ,         ,  ^    «*  ^ 

bx  \      b' 

_ — aa;+a6<y, 


12*  Prove  that  ^  +  ->2  when  a  and  h  are  bolJb  positive  or  both 
0     a 


SIMPLE  INEQUATIONS.  259 

8.  Prove  that  a^  +  6'  is  equal  to,  or  greater  than  2a6,  according  as 
a  and  h  are  equal  or  unequal. 

9.  Prove  that  a'  + 1  is  equal  to,  or  greater  than  a'  +  a,  according 
as  a  is  equal  to  1,  or  is  a  positive  quantity,  that  differs  from  1. 

10»  Prove  that  a^  +  l<a'*-fa  when  a  is  a  negative  integer,  or  an 
improper  fraction,  that  differs  from  —1. 

11.  Prove  that  «"  +  !>«'  +  «  when  a  is  a  negative  proper  fraction. 

12.  Pr 

negative. 

13(  Prove  that  j--\ — <2  when  a  and  h  are  not  both  positive  or  both 

negative. 

14,  Prove  that  a— 6>{|^a— 1^6).'  when  a>6,  and  both  are  con- 
sidered positive. 

15t  Prove  that  -r — r;,> —  when  a  and  h  are  unequaL 

a^  +  lr      a-\-h  ^ 

cc        u       \      \ 

16«  Prove  that  — +  — > — I —  when  x  and  y  are  unequal. 
y       X      X     y 

17.  Given  j    ,_  ,    ^a  ("  *^  fi^^  *^®  l^^i*  of  a;y. 

^?zs.  xy^ac-^hd. 

18.  Prove  that  a'  +  6'  +  c'>a6  +  ac4-6c  when  a,  6,  and  c  are  not  all 
equal. 

19.  Prove  that  1^3+3  V2  >  VQ  +  VV, 

ft  _J_  /J  _1_  fi  Q  Q.        Q  /T 

20.  Prove  that        ,      >^  and  <-,  -.being  the  least,  and  ^  the 

greatest  of  the  fractions,  -,  -,  and  7:. 

21.  Prove  that  xyz'y{x  +  y—z){x  +  z—y)[y-\-z—x)  when  x,  y,  and 
z  are  not  all  equal. 

22.  A  shepherd  being  asked  the  number  of  his  sheep,  replied,  that 
double  their  number  diminished  by  7  is  greater  than  29,  and  triple 
their  number  diminished  by  6  is  less  than  double  their  number  in- 
creased by  16.     What  was  the  number  of  his  sheep  ? 

Ans,  19  or  20. 


260  SIMPLE  INEQUATIONS. 

23 «  A  market  woman  has  a  number  of  oranges,  such,  that  tnple  the 
number  increased  by  2,  exceeds  double  the  number  increased  by  61 ; 
and  5  times  the  number  diminished  by  70,  is  less  than  4  times  the 
number  diminished  by  9.     How  many  oranges  has  she  ? 

Ans.  ,60. 

24,  The  sum  of  two  whole  numbers  is  25  ;  if  the  greater  be 
divided  by  the  less,  the  quotient  will  be  greater  than  3 ;  and  if  the 
less  be  divided  by  the  greater,  the  quotient  will  be  greater  than  |. 
What  are  the  numbers  ?  Ans.  20  and  5. 

25.  What  whole  number  is  that  which,  if  doubled  and  diminished 
by  6,  is  greater  than  24  ;  but,  if  tripled  and  diminished  by  6,  is  less 
than  double  the  number  increased  by  10  ? 

Ans.  There  is  no  such  whole  number. 

26.  The  sum  of  two  whole  numbers  is  32,  and,  if  the  greater  be 
divided  by  the  less,  the  quotient  will  be  less  than  5,  but  greater  than 
2.    What  are  the  numbers  ?  Ans.  24  and  8. 

27.  Twice  a  certain  number,  increased  by  7,  is  not  greater  than  19 ; 
and  thrice  the  same  number,  diminished  by  5,  is  not  less  than  13. 
What  is  the  only  number,  whether  whole  or  fractional,  that  will 
satisfy  these  conditions  ?  Ans.  6. 

28 •  Four,  added  to  five  times  a  whole  number,  is  greater  than  19, 
added  to  twice  the  number ;  and  4,  subtracted  from  five  times  the 
number  is  less  than  4  added  to  4  times  the  number.  What  is  the 
number  ?  Ans.  6  or  Y. 


Ans,  x=z4e. 


bers.  Ans»  a?=5. 

^o 


(  2x 4:<C'  x4-   6  ) 

29 1  Given  i  ^^"„    .  ,^'  r  to  find  a?  in  whole  numbers. 

(  5ic-f7>3fl;+13,  ) 

Ans,  X 
^•^^''  USS'JtfoK^^lHx;}  to  find,  in  whole  nuza. 


CHAPTER  XI 
aUADRATIC   EaUATIONS.* 

(303.)  Quadratic  Equations  are  divided  into  Pure  Quadratics 
and  Affected  Quadratics. 

PURE    QUADRATICS. 

(3 04.)  A  pure  quadratic  equation  is  one  in  which  the  unknown 
quantity  appears  in  but  one  term,  and  is  affected  by  the  exponent  2, 
or  by  a  fractional  exponent,  which,  when  reduced  to  its  lowest  terms, 

2  _4_ 

has  2  for  its  numerator ;  as,  a;' =9,  a; ^^  =4,  and  a;'  <>  =16. 

(305  •)  Every  pure  quadratic  equation  can  be  reduced  to  the 
form  a;^=a',  in  which  a^  may  represent  any  quantity,  whether  real  or 
imaginary,  positive  or  negative.  Thus,  a;f =4  is  the  same  as  (arJ^)'=4, 
which  becomes  (a;)' =4,  or  a;'' =4,  by  putting  x  for  xk, 

I 
PROBLEM. 

(306.)  To  solve  a  pure  quadratic  equation. 

SOLUTION. 

Since  every  pure  quadratic  equation  can  be  reduced  to  a;'=a',  we 
have  only  to  find  the  solution  of  this  equation. 

Taking  the  square  root  of  both  members,  we  have  x=±ia. 

We  do  not  place  the  double  sign  before  a;,  because  we  seek  only 
the  plus  value  of  a;.  . 

ANOTHER     SOLUTION. 

By  transposition,  x^^a^  becomes  a;''— a'=0,  which,  being  factored, 
gives  (a:— a)(a:  +  a)  =  0. 

It  is  evident  that  this  equation  will  be  satisfied  by  putting  either 
factor  equal  to  zero, 

*  Quadratic  equations  are  also  called  equations  of  the  second  de&bee. 


262  QUADRATIC  EQUATIONS. 

.  • .  x—a=0 
or     x-\-a=0 
which  being  solved,  give  x=a  and  x=^a. 

Scholium. — These  solutions  show.^at  a  pure  quadratic  equation 
may  be  satisfied  by  substituting  for  the  unknown  quantity  two  values 
which  are  numerically  equal  but  of  opposite  signs. 

FBOBLEM. 

(307.)  Given    a:'— 17=130— 2a;»  to  find  the  values  of  x. 

SOLUTION. 

«'— 17  =  130— 22:*, 
3a;'=147, 
x^=  49, 

a;=±7. 

EXAMPLES. 

1,  Given  4a:''  +  5=a;'  +  8  to  find  the  values  of  a?.     Ans.  a;=dbl. 
2*  Given  3x^-{-3=x^-\-6  to  fiiid  the  values  of  a;. 

Ans,  x=-±^V6. 
3.  Given  x^  +  ab=5x^  to  find  the  values  of  x.     Ans.  x=z±:^Vab, 
4«  Given  ax^-\-d=bx^+c  to  find  the  values  of  x. 


Ans,  x^=.-^a/ — ; 


/J.9     2      a;'     3 

5f  Given  -r  +  o="^  +  o  ^  ^^  *^®  values  of  x. 
4      3       8      2 


Ans,  x—±L\V\h, 

Ans,  aj=db3. 

7.  Given  4a;'— 8a;"=l  to  find  the  values  of  x,        Ans,  a;=±-J. 

'ar. 

Ans.  ar=±|V77. 


2ar     a;' 4-3 
6*  Given  — =— -—  to  find  the  values  of  x,  Ans,  a? =±3. 

3         2a; 


a;' 4- 2 
8.  Given  3a;'— 4=—— r-  to  find  the  vallies  of  a;. 
ox 


PROBLEM 


2a' 

(308.)  1.  Givena;4-V'a'+a;'=,7z=  (1)  tofindthevaluesofa;. 


> 


QUADRATIC   EQUATIOl^S.  268 


SOLUTION. 


«V^H^+aM^'=2a'  (2)  =  (1)  xVa'  +  x\ 

x^a'^x^=a''—x'  (3)  =  (2)  transposed. 

aV  +  a:*=a'-2aV  +  fl;*    (4) =(3)'. 

9x^=Sa''     _  (7).       _ 

Saj^iaf/S^  (8)=f'(7). 

PROBLE  M 

2.  Given  \/^,-\-b^-j/~-h^=h  (1)  to  find  the  values  of «. 

SOLUTION. 


i/4+6^=i/-^-6'  +  ft  (2)=(1)  transposed. 

f^      X  ^      X  

^+y=^-6'  +  264/^-6'+6'  (3)=(2)'. 
or  X  '    X' 

*«=  2hi/~—b^      ,  (4)=(8)  traiffiposed. 

\=\/^V  (6)=(4)^2i. 

^=«;_6.  (6)=(5)'. 

(7) =(6)  transposed,  ice. 


4  ~  »' 


S  («)=(^)^£ 


4a^  (9). 

256'  ^  ' 

«=±?^6  (10)  =  »^(9). 

EXAMPLES. 


1.  Given  ^^♦'^'-^'^^  to  find  x.  Am,  x=ztV2ab^b\ 

X  X  b         ' 

2.  Given  ^  =6  to  find  a?.  ^w*.  x=±-r--V2b-\-\. 

Ya'  +  x'-^x  26  +  1 


264  QtTADRATIC  EQUATION'S. 

^  3«  Given  4/^+3—4/^—3—1/-^  to  find  jr.     Ans,  z=±9V2. 
r    4  r    4  r     3 

4,  Given  V^x  +  2  —  V^x—2  =  Vx+S—  Vx—3  to  find  x. 

Ans.  a;=±5. 

5t  Given 1 =ax  to  find  x, 

x+V2—x^     x—S/2—x''  1    

Ans.  a;=±-V^a(a  +  l.) 

6«  Given =— r  to  find  a?.     -4/i5.  a:=±4. 

7.  Given  {x-\-dfz=i to  find  x. 


{x-^aY  ^^^^  a;=±(2a*  +  2a6  +  6')i 


8.  Given =0  to  find  x,  Ans.  x=:±:- . 

»•  Given  -=:=:r= =-,  to  find  x. 

ya'-x'-^Vb'  +  x'     d  

Ans.  x-±y  2{d'  +  c^)         ' 

10.  Given i^(^  +  ^+_VcL-x^/x  ^  ^^ ^^    ^^^^  ^^ ±2VS6=^. 


SIMULTANEOUS   EQUATIONS. 


PROBLEM 

x  +  yix—i/iiaib,  \  to  find  the  values  of  x 
and  y. 

SOLUTION. 

From  the  proportion,  we  get  aa;— ay = 6a; +  5y  which  becomes,  by 


(309.)  1.  Given  j;t!--^-  =  ^-} 


c 
substituting  -  for  y  as  obtained  fi'om  xy=c'y 

ac^     .        be*  .. 

ax =bx  H (1). 

X  X  ^  ^ 


{a—b)x^=c\a  +  b)  (2)=(1)  x  ar,  &c 


QUADBATIC  EQUATIONS.  265 

PROBLEM 

2.  Given   <  o     ,  ^  )J^  r  to  find  the  values  of  x  and  v. 

^'^  GLUT  I  ON. 

This  problem  may  be  solved  without  reducing  the  equations  to  the 
general  form  of  pure  quadratics. 

x^^2xy^f=.  36  (3)=(l)-(2). 

x-y^^^  (4)=V(3). 

^=±9  (5)  =  (l)-(4). 

y=±3  (6)=(2)-(4). 

EXAMPLES. 
C  3?  -I-  i/*'2/**3*l     i 

1.  Given  i       ^ ';  *  *    '   '  [•  to  find  the  values  of  x  and  y. 

^?2«.  a;==h6,  and  y=dr3. 

2*  Given  ■{    ,  .  /  f  *  *     *    '  r  to  find  the  values  of  x  and  y, 

Akis.  a:=:±9,  and  y=±6. 

3*  Given  •]    ,      aZo     '  [  to  find  the  values  of  a;  and  y, 
(  X  — y  — y,      ) 

Ans,  a;=db5,  and  y=zfc4. 

.    *Q     ,\«*      r  to  find  the  values  of  x  and  y. 
ajy+y''=126      )  ^ 

Ans,  ic=dbl5,  and  y==t6. 

C  doi? "}-  hxv  ^n  c'        ) 
5«  Given  •{  ^       '      >■  to  find  the  values  of  x  and  y. 

{  x^y  :  a; : :  m :  w,  )  ^ 


-47W. 


nc 


7ha  +  nh—mV 
c{n — m) 


mb' 


o(w-m)    /         nc 
n      y  na-{-nb— 

6.  Given  ]  ^  +  2/  •  a;;.     •    »  j.  ^^  g^^j  the  values  of  a;  and  y. 
(  a;y=6,  )  ^ 

Ans.  a;i=zfc:3,  and  y=db2. 

f  a;''  +  a:y=12,  ) 
!•  Given  •{  rt i   f  to  find  the  values  of  x  and  v, 

^W5.  a;=±2,  and  y=:±4. 

ix^  -{-y  Vx^—  9   ) 
V^4-xVxz=l^  \  *^  ^^^  *^®  ^^"^^  ^^  ^  ^^  y- 

^rw.  a;=  ±  1^  and  y=  ±4. 


266  QUESTIONS  PRODUCING  PURE  QUADRATICS. 

^    ^.         (a;'  +  a;  V^=  208,) 

9,  Given  <    ,        ,  -4-    _^  '  }•  to  find  the  values  of  a;  and  y. 

^?w.  a;=:±8,  and  y=db27. 
3         __Q .   C  to  find  the  values  of  a;  and  y. 


QUESTIONS   PRODUCING    PURE    QUADRATICS. 

QUESTION 

1.  There  are  two  numbers  in  the  proportion  of  4  to  6,  the  diflferenoe 
of  whose  squares  is  81.     What  are  the  numbers  ? 

SOLUTION. 

Let        4a?  =one  of  the  numbers, 
then       5x  =the  other  number, 
.-.  25a;'^— lear'^^Sl, 
9a;' =81, 

X  =±3, 
4a;  =±12,  one  of  the  numbers, 
bx  =±16,  the  other  number. 

QUESTION 

2.  What  two  numbers  are  those  whose  sum  is  to  the  greater  as  10 
to  7  ;  and  whose  sum  multiplied  by  the  less  produces  270  ? 

SOLUTION. 

Let         10a;  =  their  sum, 
then         7a;  =the  greater  number, 
and  3a;  =the  less  •* 

.-.      30a;'' =  270, 

a;" =9, 

X  =±3, 

7a;=±21,  the  greater  number, 

3a;=±  9,  the  less  " 

QUESTION  S . 

1.  What  two  numbers  are  those  whose  difference  is  to  the  greater 
as  2  to  9,  and  the  difference  of  whose  squares  is  128  ? 

Ans.  ±18,  and  ±14. 


QUESTIONS  PRODUCING  PURE  QUADRATICS.  267 

2.  What  three  numbers  are  those  which  are  in  the  proportion  of  ^, 
f,  and  f,  and  the  sum  of  whose  squares  is  724  ? 

^ws.  ±12,  ±16,and  d=18. 

3.  A  merchant  bought  apiece  of  cloth  for  |324  ;  and  the  number 
of  dollars  he  paid  for  a  yard  was  to  the  number  of  yards,  as  4  to  9.  How 
many  yards  did  he  buy,  and  what  was  the  price  per  yard  ? 

Ans.  27  yards,  at  $12  a  yard. 

4.  A  detachment  from  an  army  was  marching  in  regular  column 
with  6  men  more  in  depth  than  in  front.  The  front  was  afterwards 
increased  by  845  men,  and  by  this  movement,  the  detachment  was 
drawn  up  in  5  lines.     How  many  men  were  in  the  detachment  ? 

Ans.  4550. 

5.  Two  partners,  A  and  B,  divided  their  gain,  $60,  of  which  B 
took  $20.  ^'s  money  had  been  in  trade  4  months,  and  if  50  be 
divided  by  the  number  of  dollars  A  had  in,  the  quotient  will  give  the 
number  of  months  that  ^'s  money,  which  was  $100,  had  been  in 
trade.  How  much  money  had  A  in  trade,  and  how  long  had  ^'s 
been  in  trade. 

Ans.  A  had  $50,  and  B^s  had  been  in  trade  1  month. 

6.  A  and  B  invested  some  money  in  speculation.  A  disposes  of 
his  bargain  for  $11,  and  gains  as  much  per  cent,  as  B  invested;  jB's 
gain  was  $36,  and  the  gain  upon  ^'s  investment  was  4  times  as  much 
per  cent,  as  upon  B\     How  much  did  each  invest  ? 

Ans.  A  invested  $5,  and  ^  $120. 

7.  A  dog  started  in  pursuit  of  a  hare  which  was  7  rods  ahead  of 
him,  and  after  running  20  rods,  he  observed  that  the  hare  stmck 
off  at  right  angles  to  her  former  course.  He  then  changed  his  course 
so  that  he  might  overtake  her  without  another  tack.  How  far  did  the 
dog  run,  provided  he  ran  10  rods  a  minute  and  the  hare  8,  in  the 
same  time  ?  Ans.  25  rods. 

8.  What  two  numbers  are  those  whose  difference  multipliphed  by 
the  greater  produces  40,  and  by  the  less  15  ?      Ans.  ±8  and  ±3. 

9.  What  two  numbers  are  those  whose  difference  multiplied  by 
the  less  produces  42,  and  by  their  sum  113  ?      /o  J  . 

Ans.  ±13  and  ±6. 

10.  A  man  bought  a  field  whose  length  was  to  its  breadth  as  8  to 
6.  The  number  of  dollars  that  he  paid  for  1  acre  was  equal  to  the 
number  of  rods  in  the  length  of  the  field  ;  and  13  times  the  number 


268  ATTEOTED  QUADRATICS. 

of  rods  round  the  field  equaled  the  number  of  dollars  that  it  cost. 
What  was  the  length  and  breadth  of  the  field  ? 

Am.  Length,  104  rods ;  breadth,  65  rods. 

11.  A  stack  of  hay,  whose  length  is  to  its  breadth  as  5  to  4,  and 
whose  height  is  to  its  breadth  as  7  to  8,  is  worth  as  many  cents  per 
cubic  foot  as  it  is  feet  in  breadth ;  and  its  whole  value  is  224  times  as 
many  cents  as  there  are  square  feet  upon  the  bottom.  What  are  the 
dimensions  of  the  stack  ? 

Ans,  Length,  20  feet ;  breadth,  16  feet ;  and  height,  14  feet. 

12.  One  number  is  m'  times  as  much  as  another,  and  their  product 

is  n^.     What  are  the  numbers  ?  .         ,  ,    .  w 

Ans,  r±imn  and  ±— . 
m 

13.  What  two  numbers  are  those  which  are  in  the  ratio  of  8  to  6, 
and  whose  product  is  360  ?  Am.  ±24  and  ±16. 

14.  What  two  numbers  are  those  whose  sum  is  to  their  diflferenoe 
as  8  to  1,  and  the  difference  of  whose  squares  is  128  ? 

Am,  ±18  and  ±14. 


^  »>  ♦  ««  ^ 


AFFECTED    QUADRATICS. 

(310.)  An  Affected  Quadratic  Equation  is  one  which  con- 
tains the  unknown  quantity  in  but  two  terms :  its  exponent  in  one 
being  double  that  in  the  other,  and  the  least  exponent  being  either 
one,  or  a  proper  fraction  whose  numerator  is  one  ;  as, 

*5a;'  +  7a;=21;  (a— %''  +  (c4-c?)a;=e+/;  a;  +  3a;2z=8;  4a;3 +6a;3  =  13; 

Remark. — Various  plans  may  be  adopted  for  ascertaining  the  values  of  the 
unknown  quantity  in  an  affected  quadratic  equation. 

In  order  that  the  most  expeditious  mode  of  solution  may  be  adopted,  special 
regard  must  be  had  to  the  peculiarities  of  the  problem  An  aptness  in  observ- 
ing the  elements  of  a  problem  which  indicate  its  best  solution  can  only  be  ob- 
tained by  practice. 

The  student,  by  a  careftil  study  of  the  illustrative  solutions  which  follow,  will 
be  able,  it  is  hoped,  to  solve  all  the  problems  that  are  appended  to  them. 

-      1         -      1        -I 
*  The  student  should  bear  in  mind  that  y/x=x- ;   X/x^^x"^ ;   ^05=0;*, 

_2  31  2 

&c.,  and  ^ar'=a;^ ;   \fx^=^x^=x'^\   \/o^—x^^&q. 


AFFECTED  QUADRATICS.  289 

PROBLEM 

1.  Given  a;'— 2a!=— 1  (1)  to  find  the  values  of  «. 

SOLUTION', 

jr*— 2a?+l=:0  (2)=(1)  transposed. 

Taking  the  plus  value  of  zero,  «— 1=  -f  0,  or  a:=l, 
"         "    minus     "         "      a;— 1  =  — 0,  or  a;=l. 
This  solution  shows  that  x  has  two  values  which,  in  this  case^  are 
identical.    This  fact  may  be  better  illustrated  by  the  following 

SOLUTION. 

a;'-2a?4-l=0  (2). 

(a;~l)(a;-l)=0  (3) =(2)  factored. 

This  equation  may  be  satisfied  by  placing  either  of  the  binomia- 
factors  equal  to  zero.    Whence,  we  obtain  the  two  simple  equations, 

ar-l=0 

and  a?~l=0 

"Whose  solutions  give  «= 1,  and  x=  1, 

PROBLEM 

2.  Given  3^—2ax+a'=:b^  (1)  to  ^^  ^^  values  of  a?. 

SOLUTION. 

x-a=±:b  (2)=  1^(1). 

x=adtb. 
Whence,  we  find  that  x  has  two  values ;  viz.,  a-\-b  and  a— 6. 
The  same  result  may  also  be  obtained  by  the  following 

SOLUTION. 

Sincea;''—2aa;+a''=(ar— a)'',  Equation  (1)  transposed, 
becomes  (a:— a)"— 6'=0  (2) 

{x-a  +  b){x-a-b)T=0  h)=(2)  factored. 

This  equation  may  be  satisfied  by  putting  either  of  the  above  fac- 
tors equal  to  zero.     Whence,  we  obtain  the  two  simple  equations, 

x—a—b=:0 
and  «— a  +  6=0 
Whose  solutions  give  x=a^b  and  ar=a--6,  the  same  as  before. 


270  AFFECTED  QUADRATICS. 

EXAMPLES. 

]•  Given  a;'-.4a?+4=0  to  find  x,  Ans,  a;=2,  or  2. 

2*  Given  3?'  + 2a: +  1=0  to  find  x.  Ans,  a;=  — 1,  or  —1. 

3*  Given  a;''  +  4a;4-4=0  to  find  x.  Ans.  a;=  — 2,  or  —2. 

4.  Given  a;*  +  6a;  +  9=0  to  find  x,  Ans.  a;=  —  3,  or  —  3. 

5«  Given  a;'— 6a?  4- 9=0  to  find  a?.  Ans.  a:=3,  or  3. 

6«  Given  a;'— 8a; +  16=0  to  find  x.  Ans.  a?=4,  or  4. 

7.  Given  a;'+8a;  +  16=0  to  find  x.  Ans.  a;=— 4,  or  —4. 

8.  Given  ar*  + 10a; +  25=0  to  find  x.  Ans.  a;=— 5,  or  —5. 
9»  Given  a;'  — 10a;+25=0  to  find  x.  Ans.  x=5,  or  6. 

10.  Given  a;'— 12a;  +  36=0  to  find  x.  Ans.  x=e,  or  6. 

11.  Given  a;'+12a;+36=0  to  find  x.  Ans.  x=—6,  or  —6. 

12.  Given  a;' +  14a; +  49=0  to  find  x.  Ans.  x=—1,  or  —1. 

13.  Given  a;"— 14a; +49=0  to  find  x.  Ans.  x—1,  or  1 

14.  Given  a;' +  16a; +  64= 81  to  find  x.  Ans.  a;=l,  or  —IT. 

15.  Given  a;"— 18a;  +  81  =  100  to  find  x.        Ans.  a;=19,  or  —1. 

16.  Given  a;' +  20a; +  100 =64  to  find  x.     Ans.  a;=— 2,  or  —18. 

17.  Given  a;'=22a;— 121  to  find  x.  Ans.  a;=ll,  or  11. 

18.  Given  a;'— 169=24a;— 144  to  find  x.      Ans.  a;=25,  or  —1. 

19.  Given  43;"— 4a; +  1=4  to  find  x.  Ans.  a;=l|^,  or  —\. 

20.  Given  4a;'— 8a;+4=9  to  find  x.  Ans.  a;=2i,  or  — i. 

21.  Given  253;"— 20a;  +  4=16  to  find  x.        Ans.  a;=li,  or  — |. 

22.  Given  2a;''— 2a; +  1=3;'' +9  to  find  x.  Ans.  a; =4,  or  —2. 

23.  Given  4a;''— 4a; +  4=33;" +25  to  find  x.      Ans.  x=1,  or  —3. 

24.  Given  6a;'— 6a;+9=6a;''  +  l  to  find  x.  Ans.  a;=4,  or  2. 

25.  Given  9a;'— 8a; +  16  =  83?' +  25  to  find  x.     Ans.  a;=9,  or  — 1. 

26.  Given  5a;' — 1 2a;  +  9 =a;'  +  36  to  find  x.    Ans.  a;=4|,  or  —  1^. 

27.  Given  100a;'— 36a; +  4= 19a;' +  49  to  find  the  value  of  a;. 

Ans.  a;=l,  or  — f. 

28.  Given  a;'— 4a; +  1  =  — 2a; +  9  to  find  x.       Ans.  a;=4,  or  —2. 

29.  Given  4a;'  +  9a;  +  16=4  — Ya;  to  find  x.    Ans.  a;=  — l,or  —3. 


AFFECTED  QUADRATICS.  271 

30*  Given  Sic'— 6aJ  +  4=4a?'— 2a;  + 16  to  find  the  value  of  x. 

Ans.  a;=6,  or  -2. 

31.  Given  9a;'-f  12aj— 25=6a;''4-4a;— 4  to  find  the  value  oi  x, 

Ans.  a;=l^,  or  —  3|. 

32.  Given  26ar'  +  20a;— 81 = 9a;'  +  4a:— 4  to  find  the  value  of  x, 

Ans,  x=li,  or  —  2f . 

33.  Givena;"— 2aa;-f-a':^6  tofind  a;.  Ans,  x=ia^Vh. 
31*  Given  iaj'  +  ^a;- 9=— |  to  find  x.       Ans,  a;=7^,  or  —  lO^. 

4^ 
x'-Yx  +  l 

36.  Given  J^a;"+a;4-16=16a:'  to  find  x,    Ans,  a;=lJy,  or  — 1|. 

37.  Given  —  +  -  +  -=12  to  find  x,  Ans,  a;=-|±6V^3. 

^V      2arB       f^  f^         f* 

38.  Given  — +=^=0  to  find  x.  Ans,  x='^—,  ot<-, 

r       9       f  ^9       ^9 

-  Vcd 

39.  Given  cx^—2cxVd=dx''—cd  to  find  x.         Ans,  x 


i.  Given  — — ;; — 7-;  =1  to  find  x,  Ann,  x=5,  or  —3. 


Vmn 


40.  Given  ma;* + WW =2ma?Vw  +  fia;' to  find  a;.   Ans,  x: 

Vm-j^  Vn 

41.  Given  ahx^—2x{a-\-b)Vahz=z{a—hy  to  find  the  value  of  a;. 

a  +  6±i/2aN^^ 
Ans,  a;= =r= . 

Vah 

42.  Given  aa;''  +  6''+c'=a'  +  26c  +  2(6— c)a;Va  to  find  the  value  of  x, 

Ans,  x= 1^ — . 

43.  Given  a;'— a;=— :i  to  find  the  value  of  a;.       Ans.  a;=i,  or  |. 

44.  Given  a;"— 10a;=— 25  to  find  x.  Ans,  a;=5,  or  5. 

(3 1 1  •)  In  order  to  solve  the  following  problems,  the  student 
should  be  familiar  with  the  modes  of  eliminating  radicals,  and  should 
also  be  able  at  sight  to  detect  the  squares  of  binomials. 

PROBLEM 

.  1.  Given  i/^^^  +  2l/-^=b^4/-^  (1)  to  find  the  values  of  a;, 
~      X  ^  x-\-a        r    x+a  ^  ' 


272  AFFECTED  QUADRATICS. 

SOLUTION. 

'^Wh^'   (2)=(i)x4/^4:« 

The  student  might  not  be  able  to  see  that  the  left  hand  member  of 
this  equation  is  a  perfect  square,  and  not  observing  this,  the  solution 
would  be  much  more  difficult  for  him.  Let  us  endeavor  to  render 
this  fact  more  apparent.  . 


1  +-+2|/-=6»  (3)=(2)  divided. 

l  +  2-^-+-=6'  (4)=(3)  arranged. 


(1=.*)'-^ 


(i=F*y 


PROBLEM 


g/p \  VSx 1 

2.  Given  -= =1H —  to  find  the  values  of  a?. 

VSx+ 1  2 


Ans,  a; =3,  or  J-, 


SOLUTION 


Since,  Bx-l  =  (V3x  +  l)(VSx^l), 

y^x 1 

whence,    V3x — 1 = H — ^ 

2 

2f3^=4+V8^— 1, 

3ar=9, 

x=S,  , 

But  how  is  the  value  i  obtained?  By  putting  ^^33;  + 1,  which  is 
used  in  reducing  the  fraction  in  the  first  member  of  the  equation,  equal 
to  zero. 


AFFECTED  QUADRATICS.  273 

ThllS,  ^3^+1=0 

But  if  we  attempt  to  verify  the  given  equations  by  substituting  ^ 
for  the  value  of  x,  we  obtain  --=!+-. 

Is  this  a  true  equation  ?     It  seems  not ;  for  by  dropping  -  from 

both  members,  we  obtain  0=1,  an  absurdity. 

Are  we  then  to  conclude  that  a; =i  is  not  true  ? 

In  substituting  i  for  ic,  we  assumed  that  V^3a;=4-1,  whereas,  we 
leam  from  the  equation 

that  V'3V=  — 1. 
If  we  substitute,  keeping  in  mind  the  feet  that  VSx  must  be  taken 
equal  to  —1,  we  shall  obtain 

5=0,  or  0=0, 

from  which  we  leam  that  a;=i  is  a  true  value  of  a?. 
From  this  solution  we  may  infer  the  following 

PRINCIPLE. 

Ani/  expression  containing  the  unknown  quanity  that  mil  divide 
both  the  numerator  and  the  denominator  of  any  fractimi  in  the  equa- 
tion^ being  placed  equal  to  zero^  will  give  as  many  values  of  the  un- 
known quantity  of  the  given  equation  as  the  unknown  quantity  in  the 
equation  thus  formed  has  values. 

Since,  a  factor  of  the  kind  referred  to  in  this  principle  may  by  mul- 
tiplication become  common  to  both  members  of  the  equation,  we  ob- 
tain the  following 

PRINCIPLE. 

Any  expression  containing  the  unknown  quantity  that  will  divide 
both  members  of  an  equation^  being  placed  equal  to  zero,  will  form  an 
equation  of  which  every  value  of  the  unknown  quantity  will  also  be 
a  value  of  the  unknovm  quantity  in  the  given  equation. 

18 


274  AFFECTED  QUADRATICS. 

EXAMPLES. 

1,  Given    _f~   =1  -i ^^^^  to  find  the  values  of  x, 

V5X  +  S  2 

Ans.  x=5,  or  }. 

2«  Given  ^^     =c-\ — to  find  the  values  of  x. 

Vax  +  b  ^ 

Arts.  a;=-(  h  H v  I ,  or  — 

4-4-:c 

3,  Given  V64+a;'— 8a;=-===-  to  find  the  values  of  x. 

Ans.  x=3,  or  —4. 

ft  _L.  /ij 

4.  Given  \/a'' +it'—2axz=-^=:z.  to  find  the  values  of  x, 

\/a-\-x 


a(a— 1) 
ulw^.  a;=— ^— — ^,  or  —a* 


5.  Given  5Ltfifl^±?=:6  to  find  the  values  of  x. 

a-\-x 


a 


Ans,  a;=-a±^jirj-y26— 6*" 


PROBLEM 


(312.)  1.  Given  !^''+^-'J^Jl]u>&xiihevBiueso{xmiy. 

S  OLUTI ON. 

^^+2xy\-f=s'  (3)=(ir. 

4xy=4.a'  (4)=(2)x4. 

x-^2xy  +  f=s-'-^ (5)=(3)-(4). 

x-y=^Vs'-^a'      (6)=  4/5. 
2x=s±Vs^-4a'     0)={e)  +  {l). 
2y=8^Vs'^-4.a'     (8)=(l)-(6). 
a;=i(sitV«^-4a^). 
y=i(s:p|/s«-4a«). 

PROBLEM 

2.  Given   |  Jl^.'l^^  [  to  find  the  values  of  x  and  y. 


AFFECTED   QUADRATICS.  275 


SOLUTION. 


a.«+ary4-y'=7  (3)=  (2)--(l). 

a;''-2rry  +  y»=l  (4)=   (l)^ 

^xy=Q  (5)=  (3)-(4). 

xy:=2  (6)=  (6)--3. 

s?-¥2xy+f=^_  (1)=  (3) +(6). 

a;  +  y=±3  (8)=V(Y). 

2a;=4,  or— 2  (9)=  (8)  +  (l). 

2y=2,or-4  (10)=  (8)-(l). 

a;=2,or-l  (ll)=  (9)-f-2. 

y=l,or-2  (12)=(10)-j-2. 

Remabk. — Different  artifices  may  be  employed  in  eliminating  one  of  the  nn- 
known  quantities  in  the  following  equations.  The  student  should,  however,  be 
careful  not  to  find  the  value  of  one  of  the  unknown  quantities  in  one  equation, 
and  substitute  it  in  the  other,  for  an  equation  would  then  arise  which  he  is  not 
yet  prepared  to  solve. 

EXAMPLES. 

!•  Given  \  ^^  !•  to  find  the  values  of  «,  and  y. 

(      xy=2  J 

Ans,  a;=2,  or  1 ;  y=l,  or  2. 

2t  Given  \        ^~  ^     [•  to  find  the  values  of  a;,  and  y. 
{      xy=15  ) 

Ans.  x=5,  or  3  ;  y=3,  or  5. 

3.  Given  \       ^~ ^  !•  to  find  the  values  of  a:  and  y. 

(      xy=266  ) 

Ans.  a;=19,  or  14  ;  y=:14,  or  19. 

I.  Given  \    ,       «~^«  r  to  find  the  values  of  x  and  y. 
i  x^—y'=26  ) 

Ans.  x=S^  or  —  1  ;  y=l,  or  --3. 

,__  ,_    j-  to  find  the  values  of  x  and  y. 

^Sb  +  b-a' 


Ans. 


2     ^y  3a 

'36  +  6— a' 


6*  Given  •<    ,      ,     ^  f  to  find  the  values  of  x  and  y. 

Ans.  x=2,  or  1 ;  y=l,  or  2. 


276  AFFECTED  QUADRATICS. 

yfix    a_  a  f  to  find  the  values  of  a?  and  y. 

B  ,    |._-i  q  f  to  find  the  vabies  of  x  and  y. 

-47W.  x=2l,  or  8 ;  y=8,  or  21, 

-     ^_     ^  >-  to  find  the  values  of  a:  and  y, 

Ans.  x=2j  or  —1  ;  y=l,  or  —2. 

1  X        IS  — ft  1 

10*  Given  -j    „  .    „     ,  r  to  find  the  values  of  x  and  v. 


Ans.   i^=4«±i^2i3^ 

11.  Given  "{    3  ,    3     /    .    va  f  to  find  the  values  of  aj  and  v. 

^W5.  a?=2,  or  2  ;  y=2,  or  2. 

12.  Given  "S  ^     ,    ,  ~    r  to  find  the  values  of  a;  and  y, 

(      a;2^y2  =2  ) 

Ans,  ar=4,  or  1 ;  y=l,  or  4. 

13t  Given  -J  ^^  ~~^^  ~      V  to  find  the  values  of  x  and  y. 
(  a;2— ^2=26  ) 

^ri5.  a;=9,  or  1 ;  y=l,  or  9. 

lit  Given  -j  ~  ^\  to  find  the  values  of  a;  and  y. 

{x  -\-y  —b\  ^ 

Ans.  x=4,  or  1  ;  y=l,  or  4. 

15.  Given  i        ■       Z  c  f  to  find  the  values  of  x  and  y. 
(  a;   +y  — 5  ) 

^ws.  ar=4,  or  1 ;  y=l,  or  4. 

I  a;^  4-^^=4  f 

16«  Given  ■)    3       3        ,       1     ("  to  find  the  values  of  x  and  y, 

ix^+y^=(x-^-{-y^Y) 

Ans.  ar=4,  or  4 ;  y=4,  or  4. 

17«  Given  J  ^"  +2^®  —    L  to  find  the  values  of  .t  and  y. 
(      x8yT=2  ) 

Ans.  ar=256,  or  1 ;  y=l^  or  266. 


AFFECTED  QUADKATICS.  277 


18.  Given  i  ^""^y'  "^    I  to  find  the  values  of  a;  and  y. 


V  X8 — ye 
1      3  3 

\-y}  =  l  I 
\+yi  =  5  ) 


Ans.  a;=:6561,  or  1 ;  y=l,  or  6561. 

19t  Given  ^  "*'"  "~"^°  ~  ^  J-  to  find  the  values  of  x  and  y, 

Ans.  ar=256,  or  1 ;  y=l,  or  256. 

(313«)  Every  affected  quadratic  equation  can  be  reduced  to  the 
following  form : 

it'd:zAx=:±zB, 
In  which  A  and  B  may  represent  any  quantities  whatever,  whether 
real  or  imaginary,  whole  or  fi*actional. 

PROBLEM. 
To  solve  x'±:Ax=dbB, 

SOLUTION. 

First,  let  us  examine  the  case  in  which  ^  is  an  even  whole  number. 
Putting  A=:2aj  and  B=b,  and  using  the^^ws  signs,  since  the  princi- 
ple upon  which  the  solution  depends  is  the  same  as  when  A  and  B 
are  both  minus ^  or  either  one  plus  and  the  other  minus,  we  have 
a;»  +  2aa;=6.  (1) 

We  see  by  the  principles  of  binomial  squares  that  the  first  mem- 
ber of  (1)  would  be  a  perfect  square  if  it  were  increased  by  c^.    Let 
us  then  add  c^  to  the  first  member,  and,  to  preserve  the  equality,  we 
must  also  add  it  to  the  second  member.     We  then  have 
a;''  +  2aa;  +  a''=o'  +  6  (2) 

Extracting  the  square  root,  x-\-az=zrkL  i/a^  +  b 


x=—adtzVa'-\-h 
This  problem  may  also  be  solved  in  the  following  manner : 
By  transposition  (2)  becomes  {x  +  ay  —  (a'  +  6)  =  0  (3) 

Considering  a^  +  h  as  the  square  of  Va^  +  h,  the  left  hand  member 
of  (3)  is  the  difference  of  two  squares,  and  consequently  may  be  fac- 
tored thus : 

This  equation  may  be  satisfied  by  putting  either  factor  equal  to 
zero,  whence  result  the  two  simple  equations  : 
a;  +  a— VVT6=0 


and  x-\-a  +  Va^  4-  ^=0 


278  AFFECTED  QUADRATICS. 

The  first  of  these  simple  equations  gives 

And  the  second,        x=—a—  Va^  +  h,  the  same  as  before. 

EXAMPLES. 

1,  Given  a;''  +  8aj=48  to  find  x,  Ans.  .i;=4,  or  —12. 

2t  Given  ic'  +  4a:=140  to  find  x,  Ans,  a:=10,  or  —14. 

3*  Given  «'— 6a;=— 8  to  find  x»  Ans,  a;=4,  or  4-2. 

it  Given  a;'  +  8a;=33  to  find  x.  Ans.  x=S,  or  —11. 

5*  Given  a;'— 10a;=  — 21  to  find  x,  Ans.  a;=7,  or  3. 

6*  Given  aj'  +  8a;=65  to  find  x.  Ans,  x=5,  or  —13. 

7.  Given  a;'*— 2^a;=rg' tofind  a;.  Ans.  x=p±:Vp'  +  q. 

8.  Given  a;''  +  12a;=108  to  find  x.  Ans.  x=6,  or  —18. 

9.  Given  a;'—14a;= 51  to  find  a;.  Ans.  a;=l7,  or  —3. 

10.  Given  a;*— 8a; =48  to  find  x.  Ans,  a;=12,  or  —4. 

11.  Given  a;'  +  10a;=— 24  to  find  x,  Ans,  x=—4,  or  —6. 

12.  Given  a;*  +  16a?=— 55  to  find  x,  Ans,  x=—5,  or— 11. 

il  JL 

(314.)  Equations  of  the  form  a;'*±2aa;''  =  =fc6,  n  being  integral, 
are  affected  quadratics,  and  should  be  solved  as  the  preceding  ex- 
amples. 

In  such  equations,  if  we  consider,  primarily,  that  a;"  is  the  unknown 
quantity,  as  we  should  do,  the  above  equation  becomes  of  the  general 

form,  which  we  have  already  discussed;  for,  putting  y=a;'',  y'=a;", 
and  substituting  these  values,  we  get  the  equation, 

?/^±2ay=±6 
which  is  the  general  form  referred  to. 

The  student  should  observe  that,  in  some  of  the  following  exam- 
ples, a;**  =  ±Va;,  when  n  is  an  even  number,  should  be  taken  with  the 
minus  sign,  in  order  to  verify  the  equation. 

PROBLEM 

1.  Given  a;  +  8ari-=48  (1)  to  find  the  values  of  x. 


AFFECTED  QUADRATICS.  270 

SOLUTION. 

a.+  8an^+16=64  (2)=(1)  with  16  added  to  both  members, 

ar^  +  4=db8  (3)  =  l^(2). 

ici=4,  or— 12       (4) =(3)  transposed. 
a;=16,orl44       (5)=(4)'. 
The  equations  (4)  and  (5)  show,  that  in  attempting  to  verify  the 
original  equation  with  16,  that  ark  must  be  taken  equal  to  4 ;  but, 
when  attempting  to  verify  it  with  144,  ark  must  be  taken  equal  to 
—  12,  and  not  +12. 
Let  us  solve  another 

PROBLEM 

2.  Given  a?— 6a?J=— 8  (1)  to  find  the  values  of  ar. 

SOLUTION. 

a?—  6ar^  4-9  =  1  (2)  ==  (1 )  with  9  added  to  both  members. 

xi-S  =  dbl  (3)  =  |/(2). 

iri=4,  or  2  (4)  =  (3)  transposed. 

a;=16,  or4  (5)=(4)'. 

In  this  example,  in  verifying  the  values  of  a; ;  for  a;=16,  ari^  must 
be  taken  equal  to  4  ;  and  for  a;=4,  ari-  must  be  taken  equal  to  2,  and 
this  is  just  what  the  student  would  be  likely  to  do. 

We  learn  from  these  solutions  that,  in  verifying  equations,  the 

values  of  a?*,  (n  being  even,  as  in  the  above  problems,)  must  be  care- 
fully observed. 

EXAMPLES. 

1.  Givena;  +  2ari=8  to  find  ari.  Am,  xi=2,  or —4. 

2.  Given  a; +  6ari=  16  to  find  xl.  Am.  a?i=2,  or  —8. 

3.  Given  a;  +  4arj^=12  to  find  art.  Am.  ari=2,  or  —6. 

4.  Given  a;— 8ari=48  to  find  ark.  Ans.  ari=  12,  or  —4. 

5.  Given  a;4-10a;i=3  to  find  xi.  Am.  ark=  —  5dt2V1. 
6*  Given  arl4-  12art=:13  to  find  ari.  Am.  ark=l,  or  —13. 

7.  Given  a;i+14a:i=15  to  find  xi.  Am.  a:i=l,  or  —15. 

8,  Given  arf +16a;5  =17  to  find  xi.  Am.  a^=l,  or  —17. 


280  AFFECTED  QUADRATICS. 

9.  Given  a;5V4-18a;T7o=19  to  find  the  values  of  arrio, 

Ans.  arT^o=l,  or  —-19. 
10*  Given  a; 4-201^^=21  to  find  Vx.^  Ans.  Vx=\,  or  —21. 

11,  Given  V^4-22  V^=23  to  find  \/~x,     Ans.  V^=l,  or  —23. 

12.  Given  Vi  +  24  \/x=2b  to  find  X/x.      Ans.  V^=l,  or  —25. 

(3 15.)    The  solution  of   these    examples   may  be   somewhat 
■  abridged  by  omitting  the  formality  of  completing  the  square. 

PROBLEM . 

Given  a;*— 6a; +19= 13  (1)  to  find  the  values  of  ar. 

SOLUTION. 

a;*— 6a;=  — 6  (2) =(1)  transposed. 

ar»-6a;  +  9=3  (3)=(2)  with  9  added  to  both  members. 

a:-3  =  ±f3  (4)  =  4/(3). 

«= 3  ±  V^3  (5) = (4)  transposed. 
Equation  (5)  may  be  written  immediately  from  (2)  ;  since  x  is 
found  to  be  equal  to  half  the  coefficient  of  x  taken  with  a  contrary 
sign^  PLUS  or  minus  the  square  root  of  the  known  term  after  it  has 
been  increased  by  the  square  of  half  the  coefficient  of  x.  To  prove  this 
fact  to  be  general,  let  us  assume  the  four  general  equations : 

a;'  +  2aa;=6 
ar'— 2aa;=6 
a;''  +  2aa;=— 6 
a;'— 2aa;=  — 6 
A  solution  of  these  four  equations  gives  the  following  values  of  a?, 
respectively : 

x=—a±:Va^b 
x=     arbV'oM^ 


a;=— adb^/a'— 6 


x=     aiV'a'*— 6 
From  an  examination  of  these  four  equations  and  the  values  of  the 
unknown  quantity  in  each,  we  derive  the  following 

RULE. 
WTien   an   affected   qtiadratic   equation   is   reduced   to   the  form 


AFFECTED  QUADRATICS.  281 

a;'db2aa;=±6,  the  value  of  the  unknown  quantity  may  he  found  hy 
putting  it  equal  to  half  the  coefficient  of  its  first  power  taken  with  a 
contrary  sign,  plus  or  minus  the  square  root  of  the  unknoivn  term 
after  it  has  been  added  to  the  square  of  half  the  coefficient  of  the  first 
power  of  the  unknown  quantity. 

Remark. — The  student  should  apply  this  rule  in  the  solution  of  the  following 
examples  after  they  are  reduced  to  the  proper  form. 

E  X  A  M  P  L  E  Sv 

1.  Given  a;^—6a;  + 19=11  to  find  x,  Ans.  x=2,  or  4. 


2.  Given  x^  +  6bx=c^  to  find  x.  An^.  x='-Sb±i^9b^+c\ 


X     a 


3.  Given  — h -=-  to  find  the  values  of  x.    Am,  x=zldtVl  —a'. 

a    X    a 

O/u       o  2jj— —  2  

4.  Given  ~+_=ic  +  - to  find  x,  Ans,  x-2:ii2V2. 

2      dx  3 

5.  Given  a;"-!- 12a;— 16=92  to  find  x.  Ans,  x=e,  or  —18. 

6.  Given  a;"  +  6a; +  4=59  to  find  x,  Ans,  x=:5,  or  —11. 

7.  Given  a;''— 8a; +10= 19  to  find  x,  Ans.  a;=9,  or  —1. 

8.  Given  a;'— 12a;+30=3  to  find  x,  Ans,  a;=9,  or  3. 

9.  Given  a;''  +  6a;=2'7  to  find  x.  Ans,  x=S,  or  —9. 
10*  Given  8a;' +  3 2a; =3 60  to  find  x.  Ans,  x=5,  or  —9. 

11.  Given  a;''— 8a;=14  to  find  x,  Ans,  x=4±^3d, 

12.  Given  2a;' +  8a;— 20= TO  to  find  x,  Ans,  a;=:6,  or  — ». 

13.  Given  4ar»— 8a;+6=326  to  find  x,  Ans,  a;=10,  or  —8. 

14.  Given  a;+6a;i=27  to  find  x,  Ans,  aj=9,  or  81. 

15.  Given  Vx—4:  Vx=9  to  find  x,  Ans,  a;=49'7  ±1361/13. 

16.  Given  a;— 8^5=14  to  find  x,  Ans.  a;=46rt8i/30.  " 

(316*)  Every  affected  quadratic  equation  may  be  reduced  to  the 
form  ca;'±2aa;=db6,  in  which  c,  a,  and  b  are  whole  numbers  or  surds. 

The  simplest  case  of  this  general  form,  which  is  when  c=l,  has  al- 
ready been  treated  of. 

PROBLEM 

1.  Given  ca;'  +  2aa;=6  (1)  to  find  the  values  of  a;. 


282  AFFECTED  QUADEATIOS. 

SOLUTION. 

cV+2aca:=5c  (2)=(1)  X  c 

c"a:'+2aca:+a''=a'  +  6c  (3)=:(2)witha''addedtobothmember8. 

cx+a=^V^i^Tbc        (4)=|/(3). 

ca>=—a±.V^Thc  (5) =(4)  transposed. 

«= (6)=(5)-7-c. 


PROBLEM 

2.  Given  a;'— 3aJ=40  (1)  to  find  tlie  values  of  a?. 

SOLUTION. 

As  the  coeflScient  of  x  is  odd,  this  equation  is  not  of  the  requsite 
form,  but  in  all  such  cases  it  may  be  made  so  by  multiplying  by  2. 
2a;»-6a;=80  (2)=(l)x2. 

'   4a;'-12a;=:160  (3)==(2)x2. 

4»* — 1 2a;  +  9 = 1 6 9  (4)  =  (3)  with  square  completed. 

2a;~3=±13  (5)=V(4). 

2a;=16,  or  —10  (6)=(5)  transposed, 

x=8,  or  —5  ('7)  =  (6)-f-2. 

EXAMPLES. 

1,  Given  3a;' +  2a; =85  to  find  x.  Ans,  x=5^  or  —  5|. 

2,  Given  3a;'+4a;=340  to  find  x.  Ans.  a;=10,  or  —11^. 

3,  Given  5a;''  +  6a;=63  to  find  x.  Ans.  x=S,  or  — 4i. 

4,  Given  3a;'— 14a;=  — 15  to  find  x.  Ans,  x=S,  or  If, 

5,  Given  4a;''— 6a;=108  to  find  x.  Ans.  x=6,  or  — 4|. 
6*  Given  3a;'— 2a;=65  to  find  x.  Ans.  x=5j  or  — 4i. 
7,  Given  15a;''—622a;=  — 6384  to  find  a;.    ^W5.  a;=22|,  or  18|. 

m  8.  Given  — -\ 19= 15^  to  find  x.        Ans.  a;=:9,  or  —  llf. 

3       5 

^    ,              ,              118±^13724 
9.  Given  118a;— 2ia;'=20  to  find  a;.       Ans.  x= . 

10.  Given  ^a;'- 20a;=32  to  find  a;,  Ans.  a;=4,  or  — 14. 

11.  Given  5a;''  +  4a;=273  to  find  x,  Ans.  x=1,  or  —7a. 

12.  Given  afta;"— 2aj(a  +  6)Va6=(a— &)'  to  find  x.  

.     ^       a-\-h±iV2a'  +  2ly' 

Ans.   r:= , 

i^ab 


AFFECTED  QUADKATICS.  283 

(317.)  The  student,  after  solving  the  examples  in  the  last  article, 
is  presumed  to  be  ftdly  acquainted  with  the  principles  of  their  solution, 
and  is,  therefore,  prepared  to  omit  some  of  the  intermediate  equations. 
The  general  form, 

may  be  divided  into  the  four  following  equations : 

cx^  +  2ax=by 

ca^— 2aa:=6, 

cx^  +  2ax=—bf 

cx^—2ax=:  —  b, 
whose  solutions  give,  respectively,  the  following  values  for  x  : 


—a-±^a^^hc 

c 

ff  — 

a±Va:'  +  hc 

c 

^a±:Va'-bc 

c 

X- 

a±Va'--bc 

c 
A  comparison  of  these  values  with  the  equations  from  which  they 
are  derived,  gives  the  following 

E  U  L  E. 

WTien  an  affected  quadratic  equation  is  reduced  to  one  of  the  four 
forms  indicated  by  the  general  equation,  ca;''±2aa:==t6,  the  values  of 
the  unknown  quantity  may  be  found  by  putting  it  equal  to  half  the 
coefficient  of  x,  taken  with  a  contrary  sign,  plus  or  minus  the  square 
root  of  the  joroduct  of  the  known  term  by  the  coefficient  of  x^,  after 
this  product  has  been  increased  by  the  square  of  half  the  coefficient 
of  X,  and  then  dividing  the  whole  by  the  coefficient  of  x^. 

Ebmark. — In  the  following  examples,  when  reduced  to  the  proper  fonn,  the 
student  should  write  immediately  the  values  of  x,  being  guided  by  the  rule  just 
given. 

PROBLEM. 

Given  3ic'— 30=4a;  +  2  to  find  the  values  of  a?. 


SOLUTION 

2±10 


a:=- 


=4,  or  -2|. 


284  AFFECTED  QUADRATICS. 

EXAMPLES. 

1.  Given  z^—^lx—  — 3|  to  find  the  values  of  «.   Ans,  x=^\^  or  i. 

2.  Given  23?"  — 10aJ+7  =  — 6  to  find  x.  Ans,  ar=3,  or  2. 

3.  Given  3a:"  +  4a;— 7=88  to  find  x,  Ans.  x=5,  or  — 6i. 

4.  Given  11a;'— 100a;=~201  to  find  ar.         Ans,  x=3,  or  6yV. 

x^ 

5.  Given  — +  20a;=i3a;''— 80  to  find  a?.       -47W.  a;=10  or  —24. 

a;     6 

6.  Given  -+-=6^  to  find  a?.  Ans,  a;=25,  or  1. 

o     a; 

7.  Given21a;"— 1616a;=--20'748tofinda;.  .4?wf.  a:=60|,or  16f. 

o    ^.        18a;»     18078aJ         ^^^^  ^    ^  ^ 

8.  Given  -— •+ — — — =—4728  to  find  x, 

o  00 

-47W.  a;=— 251^,  or  —  52, 

9.  Given  4a:'— 9a;=5a;'— 265J— Sa;  to  find  the  values  of  x, 

Ans,  151,  or  —16^. 
10.  Given  {U^--9cd^)a^  +  {4a''c'  +  ^abd')x—-(ac'+bdy  to  find 

ac'  +  bd^ 


the  values  of  x,  Ans.  x=z- 


11*  Given  6a;''  +  2a;=14  to  find  x,  Ans,  x= 


2adzSdVc 

-ld=i^"85 


6 

12.  Given32a'""c*^' +4a'^''c'*-*(ac«— 2)a:=a'c"+"a;»    to  find  tlrff 

o^m — * 

values  of  x,  Ans,  a:=4a'^^  or -—. 

c" 

(3 1 8.)  Let  us  now  examine  the  general  equation  ca;'±aa;=db6,  in 
which  a  is  an  odd  number.  It  is  evident  that  the  rule  given  in  the 
last  article  is  also  applicable  here,  but  in  order  to  avoid  multiplying 
the  equation  by  2,  or  in  case  this  should  not  be  done,  to  avoid  firao- 
tions,  we  seek  another  mode  of  solution. 

PROBLEM. 

Given  cx^+ax^zb  to  find  the  values  of  a?. 

SOLUTION. 

If  we  multiply  this  equation  by  4c,  we  shall  have 
4c  V + 4aca: = 45c, 
4cV+4gca;  +  ct'=«'  +  45c, 

2ca;  +  a=dbVa'  +  46c, 
2ca:=  —a±:  Va''  +  46c, 
-a±|/a'  +  46c 

'= 2i — • 


AFFECTED  QUADEATICS.  285 

This  mode  of  solution  is  found  in  the  Bija  Ganita,  a  Hindoo  Treat- 
ise on  Algebra,  which  has  been  translated  by  Mr.  Colebrooke. 

PROBLEM. 

Given  2a;"— 5fl?=ll'7  (1)  to  find  the  values  of  x, 

SOLUTION. 

16a;'— 40a;=936         (2)=(1)  X  8. 
16a;»—40a;  + 25=961. 
4a;— 6  =  db31. 

4a;=6±31=36,or— 26. 
a;=9,  or  —6^. 

EXAMPLES. 

1,  Given  a;*— 84=:Ja;  to  find  x.  Ans,  a;=6,  or  — 6|. 

2.  Given  a;'4-3a;=72  to  find  x,  Ans,  x= , 

3*  Given  5x^-{-x=4  to  find  x.  Ans,  a;=|,  or  —1. 

4*  Given  2a;"— a;=21  to  find  a;.  Ans,  a;=3|,  or  —3. 

5*  Given  ax^—oxz^c  to  find  x,  Ans,  x= . 

2a 


p-±:  Yp'  4-  4(7 

6.  Given  ar'-f^a;= g' to  find  «.  Ans,  x=—±- — -£- ±, 

7.  Given  3a;'4-6a;=42  to  find  x,  Ans,  a;=3,  or  —  4|. 

8.  Given  a;"  4- 6a; +  4= 22— a;  to  find  x,  Ans,  a;=2,  or  —9. 

9.  Given  a;"— 5fa;=18  to  find  x,  Ans,  a;=8,or  —2\, 

10.  Given  a;"— 3a;=10  to  find  x,  Ans.  x=5,  or  —2. 

X 

11.  Given  6a;"— -=78  to  find  a;.  Ans.  a;=4,  or  — 3^^. 

2 

12.  Given  4a;3 +a;«  =39  to  find  x.  Ans.  a;=729,  ot(-^\  , 

(319*)  This  mode  of  solution  may  be  abridged  by  omitting  the 
intermediate  equations.  The  relation  of  the  unknown  quantity  to  the 
known  terms  of  the  general  equation  ca;'±aa;=db6,  may  be  observed 
by  solving  the  following  equations : 

cx^-\-ax=bj 

cx'*—ax=b, 

cx^-\-ax=:  —  b, 

cx^—ax=i^b. 


286  AFFECTED  QUADRATICS. 

The  values  of  a;  in  these  four  equations  are 

—a±:VaF+Abc 

x= , 

2c  ' 


a±:Va'-\-4bc 

''= ¥c ' 


—a±:Va'—4:hc 


azhVa'—4:hc 

aj= . 

2c 

By  a  comparison  of  these  values  with  the  equations  from  which 

they  are  derived,  we  obtain  the  following 

RULE. 
When  an  affected  quadratic  equation  is  reduced  to  one  of  the  forms 
indicated  by  the  general  equation 

ca:'±aaj=±6, 
the  value  of  the  unknown  quantity  may  he  found  by  putting  it  equal 
to  the  coefficient  of  ar,  taken  with  a  contrary  sign  plus  or  minus  the 
square  root  of  the  square  of  the  coefficient  of  x,  after  it  has  been  added 
to  four  times  the  product  of  the  coefficient  of  x^  by  the  known  term, 
and  dividing  the  whole  by  twice  the  coefficient  of  x^, 

FROB  LEM.    " 

Given  Sx^—^x=  —34  to  find  the  values  of  x, 

SOLUTION. 


«= . 

16 

EXAMPLES. 

1,  Given  ea;"— a;=92  to  find  x,  Ans.  fl;=4,  or  — 3|. 

2,  Given  8x^—1x=lQ5  to  find  x.  Ans.  aj=5,  or  — 4|. 

3,  Given  3rc'— 3a;4-6=6^  to  find  x.  Ans.  x=^,  or  |. 

4,  Given  Ufa;— 3^0;'=— 411  to  find  x.  Ans.  x=—2\,  or  5^. 

5,  Given  9ia:'— 90ia;=  — 195  to  find  x.  Ans.  x=z6^,  or  3^. 

d  h 

6«  Given  adx—acx* ^zbcx—bd  to  find  x.         Ans.  a;=-,  or  — . 

c  a 


m       r^'  /  ,S     Q  ««^         .       /.      ,  A  C±lVc^  +  4:a^ 

7.  Given  ia-\-b)cir—cx-\ ^  to  find  a?.  Atis.  x^=.—~. — — . 


AFFECTED  QUADEATIOS.  287 


8.  Given  1/9a;4-4=3a:to  find  a;.  Am,  a;=lJ,or— i. 

9*  Given  ax^—hx  +  c^ca;"  4-  2c  to  find  the  value  of  x. 


6=b|/6^  +  4c(a-c) 

Ans.  x= -, ^^ \ 

2(a— c) 

10.  Given  ^^  +  -^~(a-6)(2c+ac?)^=(a4-6)-|'-(a»-6>'tofind 
the  values  of  a?.  Ans.  x=—- — — -,  or 


d{a-^hy      d{a-h)' 

11.  Given  aoa?"  H =- = to  find  the  values  of  x, 

c  c  c 

.  '2a— h  3a+26 

Ans.  x= .  or , 

ac  he 

3a;»     21a;— 2*7782 

T"^  12 

of  a?.  *  -4ws.  aj=— 46, or  24J. 

(320*)  It  is  frequently  advisable  to  consider  several  tenns  as  one 
in  the  solution  of  afieoted  quadratics  involving  radicals. 

PROBLEM. 


12.  Given  80a;  +  -^  + r:^ =  18591— 3a;»  to  find  the  values 


Given  Va;-f  12+\/a?+12=:6  to  find  the  values  of  a?. 


SOLUTION 


If  we  put  y=X/x  -h  12  ,  y'  will  equal  j^x  + 12. 

y= — -—=2,  or -8, 


and     y^—  Vx  + 12=4,  or  9, 
aj  +  12  =  16,  or81, 
a;=4,  or  69. 
In  verifying  the  value  a;=69,  we  must  take,  as  the  solution  indicates, 
y=V«Tl2  =  — 3. 

EXAMPLES. 


1,  Given  ^/aj  +  lO— Va;  +  10=2  to  find  x.      Ans.  a;=6,  or  —9. 


2.  Given4/a;  +  2H-Va;  +  21  =  12tofinda;.   ^tw.  ar=60,or  235. 


3.  Given  4/20:4-6  4- V2a;+6 =6  to  find  x.      Ans.  x=5,  or  3*7^. 

4.  Given  a; 4- 1/^+6= 2 4- 3  VW^  to  find  a;.  Ans.  a?=10,  or  —2. 


5.  Given  a;4-5=V'a; 4-54-6  to  find  x.  Ans.  a;=4,  or  —  1. 


288  AFFECTED  QtJADRATICS. 

6.  Given  a;+16~7iV+T6=10— 4  V^+16  to  find  x, 

An^,  a;=9,  or  —12. 


7.  Given  ^x  +  a-\-b  Vx+a=:2b''  to  find  of  x. 

Ans,  x—h^—ay  or  166*— a. 

(321.)  It  is  sometimes  advisable  to  complete  the  square  without 
reducing  the  equation  to  any  of  the  forms  given  above4 


PROBLEM. 

Am?     Sx 
Given  —  +—=8  to  find  the  values  of  x, 

ol        lo 


SOLUTION. 

Adding  1  to  both  members  of  the  equation,  and  we  have 
4a;"     8a;     ^     ^ 
81  +18  +  ^=^- 
2,x 

y=2,or-4. 

2a;=18,  or  —36. 
a;=9,  or  —18. 

EXAMPLES. 

1.  Given  _— _4.5=0  to  find  x.         Ans,  a;=9(l±2i^— 1). 

a;"     4a;      1 

2.  Given  — _— +  _=o  to  find  a;.  Ans.  x=S,  or  1. 

«    ^.         dx*      16a; 

3.  Given  — +  —  +  4=0  to  find  x,    Ans,  a;=  — IJ,  or  — 5f 

x^      223/ 

4.  Given  — — _=_ 32  to  find  a;.  Ans,  a;=152,  or  76. 

5.  Given  ^  -  4aa;  +  36''=0  to  find  x,  Ans,  a;=— ,  or  -. 

^  a        a 

4a;* 
6*  Given  — — 4a;=:7  to  find  x,  Ans.  a;=10i,  or  —1^. 

^    _.        aV      Sbix     126*  ^   ,  ,  6bi       2bi 

7,  Given  -^ ^  +  —^=0  to  find  a;.    Ans,  a;=-^,  or  — . 

0  a  a  a*  ^        n* 


AFFECTED   QUADRATICS.  289 

8.  Given  —-+—=61  to  find  x.  Am.  x=1^  or  — 11|. 

PROBLEM. 

(322.)  Given  x^+x=b  to  find  tlie  value  of  x,  b  being  the  pro- 
duct of  two  consecutive  whole  numbers. 

SOL  UTION. 
By  the  conditions,  we  have  by  putting  a  equal  to  the  least  of  the 
consecutive  numbers  .  x^  i-x=b=a{a-\-l) 

x'  +  x=a^-\-a. 
A  bare  inspection  of  the  last  equation  shows  that  one  value  of  a;  is 
a,  but  to  get  both,  we  complete  the  square 

x''  +  x  +  l=a^  +  a-\-l, 
.  aj  +  i=±(a4-i) 

x=a,  or  —  (a  +  1). 
From  which  we  observe  that  in  an  equation  of  the  form 
x^-]-x=b=a{a  +  l)j 
X  has  a  positive  and  a  negative  value,  the  positive  value  being  equal  to 
the  least  of  the  consecutive  numbers,  and  the  negative  one  equal  to 
the  other.     If  the  equation  were  a;"— a;=6=a(a+l),  the  values  of  x 
would  be  found  to  be  the  same  with  opposite  signs,  namely, 
a;=— a,  ora  +  1. 
Scholium. — Since,  a^-\-a  +  \=a{a-\-\)-\-\^vTQ  conclude  that  the 
product  of  any  two  positive  consecutive  numbers  increased  by  J  is  a 
perfect  square,  therefore, 

6i=(2i)». 
12i=(3i)'. 
20J=(H)'. 
&c.       &c. 

PROBLEM. 

Given  a;'  -fa?=20  to  find  x. 

SOLUTION. 

ar»-fa;  +  i=20J. 

a;=4,  or  -—6. 
We  might  write  immediately  the  value  of  x  thus 
a;=-i±:4^=4,or-6, 
or  decide  its  values  by  the  principle  mentioned  above. 

19 


290  AFFECTED  QUADBATICS. 

BX  AMPLE  9. 

1«  Given  a;''  +  «=2  to  find  the  value  of  x.      Ans,  a;=l,  or  —2, 

3»  Given  x^^x=::6  to  find  the  value  of  x.      Ans.  ic=— 2,  or  3, 

3*  Given  x"^ -\-x-=\2  to  find  the  value  of  ar^    Ans,  a;=3,  or  —4. 

4*  Given  x'^—x^:z20  to  find  the  value  of  x.      Avis.  a;=— 4,  or  5. 

H,  Given  ar'  +  a;=:30  to  find  the  value  of  ar,     Ans.  x=5,  or  —6. 

6*  Given  a;'— ar=42  to  find  the  value  of  ar.     Ans.  x=:'—6,  or  7, 

7.  Given  x''-\-x=56  to  find  the  value  of  ar.     Ans.  a; =7,  or  —8. 

8*  Given  a;'— a;=V2  to  find  the  value  of  ar.     Ans.  x=—8,  or  9. 

9*  Given  a;'  +  a?=90  to  find  the  value  of  a?.    Ans.  x=9,  or  — 10, 

10«  Given  a;"— a;=110  tofindthe  valueof  a;.  Ans.  x=  —  10,ot  11. 

11,  Given  ar*  +  a;=132  to  find  the  value  of  x.  Ans.  ar=:ll,  or  —12. 

12.  Given  a;''— a;=306  to  find  the  value  of  a:.  Ans.  a;=  — 1*7,  or  18. 

(3!23.)  The  student  is  not  always  limited  to  the  modes  of  solu- 
tion which  have  already  been  given, 

P  R  O  B  L.E  M  . 
Given  a3C^-\-hx=c  to  find  x. 

SOLUTION. 

Any  two  terms  can  be  made  a  square  by  adding  to  them  the  square 
of  the  quotient  arising  from  dividing  one  of  the  terms  by  twice  the 
square  root  of  the  other.     Hence,  ax^  4-  hx  will  become  a  square,  if 

,,        .     ,  .    hx  h       ^.  ^  .    b' 

we  add  to  it  the  square  of =:,  or which  is  -— . 

2xVa         2V'a  *« 

Adding  —  to  both  sides  of  the  given  equation,  we  have 

,     ^        b'  b'     4:ac  +  b^ 

4a  4a         4a 


r^x_      .  ,  /-        *         ,y4ac-hb' 

Extracting  square  root,  we  have  xva-i ==± = — . 

2Va  2Va 

2aX'hb=±:V4ac-\-b\ 


2aa;=  —  b±:V4ac  +  b\ 


_'-'b±V4ac  +  b^ 
~~  2a  ' 


AFFECTED  QUADRATICS.  291 

Remark. —  ax^  +  hx  will  become  a  square  when  —r-  is  added  to  it. 

a^x^        a^x^ 
That  is,6a;+aa;^  +  — ^,  or  —-^-{-ax^-^hxh  a  perfect  square.  So  also, 

will  aar'+fta?  become  a  square  when  2xVahx  is  added  to  it.  2xVahx 
is  obtained  by  considering  ax^^  and  hx  as  the  first  and  last  term  of  the 
square,  and  finding  the  middle  term  which  is  equal  to  twice  the  pro- 
duct of  the  square  roots  of  the  other  two.  Hence,  two  terms  being 
given,  we  can  find  three  terms,  any  one  of  which  being  added  to  the 
given  terms  will  produce  a  square.  The  first  one  is  the  only  available 
one  in  the  solution  of  quadratics. 

EXAMPLES. 

!•  Given  3a;'4-4a;=7  to  find  x.  Ans,  a;=l,  or  —  2i. 

2«  Given  6ar*  +  a?=22  to  find  x.  Ans,  x=2j  or  — 2J. 

(324«)  Problems  of  a  special  character  may  sometimes  be  solved 
by  modes  different  firom  any  that  have  been  given.  A  few  are  given 
a&  a  matter  of  curiosity. 

PROBLEM. 

Given  3a;"  +  4a;=  — 1  (1)  to  find  the  value  of  x, 

SOLUTION. 

3a;'  4-  4x  + 1 = 0  (2) = (1 )  transposed. 

K  the  coeflScient  of  x^  were  4  instead  of  3,  the  first  member  of  this 
equation  would  be  a  perfect  square.  In  order  to  render  it  so,  let  us 
add  x^  to  both  sides,  and  we  have 

4a;'  +  4a;+l=a;'. 
2a;+l  =  ±a?. 
2a;if:a;=:  — 1. 
a;or3a;=— 1. 

ar=-l,  or-f 

EXAMPLES. 

1,  Given  3a;'  +  8a;=  —4  to  find  x,  Ans,  a;=--2,  or  — f. 

2*  Given  16a;'4-8a;— —  1  to  find  x.  Ans.  a;=— i,  or  —J. 

3«  Given  1 53;"  + 16a:  =—4   to  find  a;.  Ans.  a;=— |,  or—f. 

4*  Given  12a;'H-8ar=  — 1  to  find  x,  Ans.  x=—^,  or  —}. 

5«  Given  12a;' 4- 16a;  =—4  to  find  x.  Ans.  a;=  — 1,  or  —^, 
6*  Given  8a;'— 12a;  =—4   to  find  a;.  Ans.  a;=l,  or  |. 


292  AFFECTED  QUADRATICS. 

(325.)  The  student  should  observe  carefully  the  solution  here 

given  of  the  equation  3a:''-+-10a;=— 8,  not  because  the  plan  given  is 

the  best,  but  that  he  may  be  able  to  solve  in  a  similar  manner  the 

examples  which  follow,  which  are  intended  to  give  him  a  foretaste  of 

I  a  principle  which  is  employed  in  the  solution  of  cubic  equations. 

PROBLEM. 

Given  Saj'  +  lOa;:::— 8  to  find  the  value  of  a?. 

SOLUTION. 

3a;'  +  10a;+8=0         (1). 
Adding  1  to  both  members  of  (1),  makes  the  last  term  of  the  first 
member  a  square. 

Thus,     3a;''  +  10aJ4-9=l         (2). 
Adding  now  «*  to  both  members  of  (2),  makes  the  first  term  of  the 
first  member  a  square. 

Thus,     4ar'  +  10a;  +  9=a;''  +  l. 
By  examining  the  first  member,  we  see  that  it  would  be  a  perfect 
square  if  the  middle  term  were  12a;  instead  of  10a:.    But  we  can  make 
it  12ar  by  adding  2a;  to  both  members,  which  being  done,  not  only 
renders  the  first  member  a  perfect  square,  but  also  the  second  one. 
Thus,     4a;»  +  12a;+9=a;»+2a;+l, 
2a;+3  =  ±(a;  +  l), 
a;=-2,-H. 

EXAMPLES. 

1*  Given  3a;'  +  8a;= — 5  to  find  x.  Am,  a;=  —  1,  or  —  If. 

2.  Given  8a;'  +  10a;=— 3  to  find  x.  Am,  a;=— ^,  or  — f. 

3*  Given  6a;''  +  8a;=  — 3  to  find  x.  Ans.  a;=— |,  or  —1. 

4,  Given  3a;''-M8a;=  — 15  to  find  x,  Ans.  x=  —  l,  or  —5. 

5.  Given  12a;''— 3 2a;  =—5  to  find  a;.  Am,  x=2ly  or  }. 
0*  Given  '7a;''— 18a;  =  —  8  to  find  x.  Am,  a; =2,  or  4. 
7.  Given  15a;'— 16a;='7  to  find  x,  Ans,  x=—i,  or  If. 
8i  Given  15a;'— 32a;=7  to  find  x,  *  Am.  a;=2i,  or  —J. 
9.  Given  9a:'  +  22a;=16  to  find  x.  Am.  a;=  — 3,  or  f. 

10.  Given  16a;'4-8a;=8  to  find  x.  Am.  a;=  — 1,  or  1. 

11.  Given  24ar'— 2a;=15  to  find  x.  Am.  a;=— f,  or  f. 

12.  Given  20a;'  +  10a;=24  to  find  x.  Am.  a;=  —  1  i,  or  f . 


AFFECTED  QUADRATICS. 


MISCELLANEOUS   EXAMPLES. 

1,  Given  -r — l=a?+ll  to  find  x,  Ans.  a;=12,  or  —6. 

6 

6        2 

2,  Given  — -—-\ — =3  to  find  x.  Am,  x=:2,  or  — ^. 

x+1     X 

3«  Given  6aJH =44  to  find  x,  Ans,  a:='7,  or  f. 

.    ^.        16     100— 9a;     „  ,    «   ,  .  .        «  , 

4*  Given —5 — =3  to  find  x,  Ans,  «=4,  or  2^^^. 

X  4:X 

d — X       X        b 

5«  Given 1 =- to  find  a?. 

«;       a-^x     c 


^^'  ^=4^^5(2^/^'-4 


-     -.       i^4a;  +  2     4-Va;  ^    .    ,  '  .        «^ 

6*  Given == =r-  to  find  x.  Ans,  a;=4,  or  -y. 

4  +V^ar         f'a; 

V^a'aj  +  ft     a— 1/a; 

7«  Given r-= =-  to  find  x, 

a+Vx         Vx 


/-6±V'4a'  +  4a''+6V 

Ans.  x^=\ —. -T I 

\  2(aH-l)  / 

8,  Given  (ya?  +  6)(t'a:  +  12)=:12  to  find  x. 


Ans.  a;=4,  or  —21. 
2 
x—8  6 

2      3^ 


^    ^.        8— a;     2a;— 11     a;— 2^    .   ,  .  «        , 

9,  Given  — ZI^~~T~  ^'  ^*  ^~  '  ^^  ^* 

10.  Given r-=22|  to  find  x,  Ans,  a;=49,  or  ^^ 


_        21 

11 .  Given  f/2a;  + 1  +  2i^a;=  -  to  find  x, 

V2x  +  l 

Ans.  a; =4,  or  —26. 

12.  Given  V^+4^ip^=6V«  to  find  x.  Ans,  a;=2,  or  —3. 

13.  GiveniV— 21^^— a;=0  to  find  x.  Ans.  x=4,  or  1. 
II,  Given  Vx^—a^=x—b  to  find  x, 

Ans.  a;=:^dr-i^V48a'6-126*. 

3 1/^__  2      1 

15.  Given  ^^ = — to  find  x.  Ans.  a;  =3 49,  or  25. 

x—5       20 


294:  AFFECTED  QUADRATICS. 


16,  Given  x+V5x-\-lQ=^S  to  find  x.  Ans.  a;=18,  or  3. 


17.  Given  a;4Yl0a;+6=9  to  find  x.  Am.  a;=25,  or  8. 

_______  ___        '7/»  _1_   K/*; 

18.  Given  2Vx-'a  +  W2x= to  find  x. 

Vx-a. 

Ans,  aj=9a,  or  — a, 

10,  Given  3a:*— a:=140  to  find  x.  Am,  x=1^  or  —  6|. 

1x 
30*  Given  5a^+-— =1x^—51  to  find  x.         Am.  x=6,  or  —4^. 

4/|. 4 

21.  Given  2a;' — =1x  to  find  x.  Am,  a;=4,  or  i, 

3 

22,  Given  3a:'-l'7a;=2ii;''  +  84  to  find  x.       Am,  a;=21,  or  -4. 
2l3t     Given  re'— a;  +  3=45  to  find  x.  Am,  a;=7,  or  —6. 

24«  Given  4a; 7=  14  to  find  x.  Am.  a;=4,  or  —  li. 

a!  +  l 

^^    ^.        10     14— 2a:     22  ^    _   ^  . 

25«  Given = — =— -  to  find  x.  Am,  x=3.  or  l\^, 

a:  a:'  9 

26.  Given  6a:'— 4a;+ 3  =  159  to  find  x.  Am.  x=6,  or  —  5|. 

J221 4a; 

27i  Given  3a: =2  to  find  x,     Ans,  a:=19,  or  — 19|. 

X 

28*  Given =— -—  to  find  x.         Am,  x=6.  or  I, 

2  a:— 3  6 

Qa. g  Q/j. g 

29t  Given  5x =2x-\ to  find  x.  Am,  a:=4,  or  —1. 

a;— 3  2 

-I  OQ  __  0„ 

30i  Given  3a: =29  to  find  x.       Ans,  a:=13,  or  —4^. 

x  . 

2a:'     4a: 
31*  Given  16 = h  V|  to  find  x.  Am,  x=S,  or  — 4i. 

3        5        *  7  5 

3ic— -4  a:— 2 

32.  Given -  +  1  =  10 -—  to  find  x.       Am,  a;=12,  or  6. 

a:— 4  2 

33.  Given  — — -= — -- —  to  find  x, 

5  x—6  10 


«.    ^.  3a:— 10     „     6a;'-40       ^  ^ 

34.  Given  3a: — - — -—=2-}-— —  to  find  x 

9— 2a:  2a:— 1 


Ans,  a: =36,  or  12. 

V, 

Am,  a?=ll^,  or  4. 


AFFECTED   QUADEATICS.  295 

^w«.  a;=l,  or  —  a«. 

90       2Y         90 

S6«  Given = to  find  a:.  -4?^.  ir=4,  or  — 1|. 

ar      a;  +  2     x-\-\ 

1  19 

37.  Given  -^ — --  +-^ — -  =7—  to  find  ic.      Am,  a?=4,  or  —  3f. 
ic'— 3a;     a;' +  4a;     8a; 

y>* 10a;' +  1 

38*  Given  — - — — =aj— 3  to  find  ar.        Am.  a;=l,  or  --2a 

a;''— 6a; +  9 

39t  Given  t-^ — y-^^^—l^-^  to  find  a?.  Am.  x—b,  or  2. 

7— a;        a; 

JA    fv        '7~12a;     a;      8a;  +  110  ^    „    , 
40*  Given — =^r =r —  to  find  x. 

xli        Vx        Vx' 

Ans.  a;=9,  or  —13. 

--     ^.        a—x  .     a;        6  ^    „    -  .  2ac 

41.  Given i =- to  find  a;.     Am.  x=. 


X        a—x     c  1c  +  h±VV'—A:C^ 


-_       __ 2j/a;'  +  60a;'  +  9a;  +  540  +  89      ^    , 

42«  Giveni/a;  +  60+1/a;'  +  9= _1_  to  find  a?. 

Vx-irQO^-Vx'  +  d 

Am,  a;=4,  or  —5. 

.-     _.        123  4-41^^^     "lOVx  +  Ax  2x^  ^    .    , 

43*  Given = = = ~  to  find  x. 

bVx'-x  S-^Vx       (5Vx—x){S  —  Vx) 

Ans,  a;=20|^,  or  3. 

X                      X  b 

44*  Given  -^^ H — := =-=  to  find  x, 

Vx-\-Va—x    Vx—^a—x    Vx 


b±Vb'-2c^ 

Ans.  x= , 

2 

45*  Given hl0a;=i95  to  find  jc    Am.  x=1,  or  —6, 

X 

46.  Given  — — 14-20^=:42|  to  find  x.         Am.  x—7,  or  — 6|. 

.„     ^.  2X+VX      ^,         ^2X—VX^    rt    :i  A  .0, 

47.  Given ^=3y^j— 3. =  to  find  x.  Am.  x=4^  or  3  Jg-, 

2x-^x  2x-\-Vx 

Ao    ^.        54-9i/^     2Sx—4:6Vx  1x*-Sx  +  A-  ^  ^ 

48.  Given =:= = 1 ir nr  to  find  X. 

x-^2Sfx        6  +  4/aj         {x^2^x){^^-S/x) 

Ans.  a;=5,  or  —  2^^. 


296  AFFECTED  QUADRATICS. 

49.  Given ^  = ,r— + =-  to  find  x, 

8-34/a;       4+4/jB         {8-SVx){4  +  t/a;) 

-4w5.  x=93j  or  Y. 

/:a    n-        i^—Vx+1       5   ^     _    , 

DO.  Given =r^=—.  to  find  a;.  ^ws.  a;=8,  or  — f. 


51.  Given  {V4x  +  5)(V1x+l)=80  to  find  x. 

Ans.  x=5,  or  —JaV-- 

^-    ^.        15a;— 5      2         - 

52.  Given =4-— =31/a;  to  find  x,  Ans.  a;=4,  or  I, 

l  +  5Vx    Vx 


53.  Given  9a*5V-6a^6''a:=6''  to  find  ar.      Ans,  x=z^^^^^^^. 


54.  Given3^112— 8a;=19+V3a;  +  7  tofindar. 

An^.  .r=6,  or  VA'-' 

55.  Given  V2x+1  +V8x—\8=V1x  +  \  to  find  a;. 

Ans.  a;=9,  or  —  3f. 


5o«  Given  -y^ 1 — 5=0  to  find  x.    Ans.  x=- k 


c       c*  a  c 


57.  Given  x+  Vx :  x-~Vx::3Vx  +  e  :  2Vx  to  find  x. 

Ans.  ar=9,  or  4. 


58.  Given  ♦/(4  +  ar)(5--ar)=:2ar— 10  to  find  x.     Ans.  x=5,  or  3^. 

59.  Given  a;'— a;— 40  =  170  to  find  x.  Ans.  a;=15,  or  —14. 

^3. 3 

60.  Given  x-] — =8  to  find  x,  Ans.  a:='7,  or  9}. 

61.  Given  2V^+3V^=2  to  find  x.  Ans.  x=l,  or  -8. 

62.  Given  4a;= f-46  to  find  x.  Ans.  a;=12,  or  — #, 

X  '           *' 

63.  Given 1-^— — =— -  to  find  x.  Ans,  x=*l.  or  —3. 

7— a;     *l-\-x     10 

M*^     r,'        3ar+6     3a;— 5     135       ^    , 


65.  Given  ♦/(a;— l)(a;-2)  + t/(a;-3)(a;— 4)=|/2  to  find  a;. 

Ans.  a;=3,  or  2. 


AFFECTED  QUADRATICS.  297 

66.  Given  ,    ,  ^._  +—^=——-^  to  find  x. 

Ans.  x=6,  or  — f . 

67.  Given  '^\/^-^-\/i  +45=1^10^+56  to  find  x. 

Ans.  ^=20,  or  -VsTVeV/- 

68.  Given  aJV  +  (l+c)6f;|/c  +  c6V=[6'(?t/c  +  (a6  +  c)^(l+c)]a;  to 

6c?1/c         1+c 
find  X,  Ans,  x=  or  -^^. 

ao  +  c  0 

a;  Y 

69.  Given r:r=r to  find  x.  Ans.  z=14,  or  —10. 

a;  +  60     3a;— 5  ' 

70.  Given f-— =13  to  find  x.  Ans.  x=9,  or  lyV 

x—5      X 

Sx  20 

71.  Given -— 6=—  to  find  x.  Ans.  a;=10,  or  — |. 

3/ +  2  oiC 

72.  Given -= tt:  —  ^  to  ^^  ^'  ^^*-  ^=^i  or  5. 

a;  ~f"  o     X  -p  xU 

73.  Given  — =:*— — r-  +  l  to  find  x.  Ans.  x=6l},  or  4A. 

0/p_i_  3         2a; 

74.  Given  — — -—— — ^  — 6^  to  find  x.     Ans.  a;=13|a.,  or  8. 

lU  —  X       JiO  —  oX 

'         n-        25a;  +  180       40a;       3^    ^    , 

75.  Given  -— r^=^ r— ;r  to  find  x. 

10a;— 81      5a;— 8     5 

Ans.  a;=14|,  or  /^. 

^-     _.  18  +  a;       20a;  +  9  65       ^     _    , 

76.  Given  -j- r=r-^ — ^"tt;^ x  to  ^^^  ^• 

6(3— a;)     19  — 7.r     4(3— a;) 

Ans.  x=1^^j,  or  2|. 
_    ^,.        5a  +  10a6\     /5i/M^^(l  +  26Vv/c\      cc^ 

,    «    ,  .  (3-an|/a  +  6        Sb'cdVc 

to  find  a;.  ^ri«.  x=^    ..    '  ^„.    ,  or . 

a6(l  +  26')    '  5a 


»»»■♦..» 


AFFECTED    QUADRATICS    INVOLVING    TWO 
UNKNOWN    QUANTITIES. 

(326.)  In  solving  equations  of  this  character  the  usual  plans  of 
elimination  may  be  employed.  The  student  should  adopt  that  plan 
which  seems  to  be  best  for  the  example  under  consideration. 


298  AFFECTED  QUADRATICS. 

EXAMPLES. 

1,  Given   I      •'*^y_  8    /   to  find  the  values  of  a;  and  y. 

.]  yr=3,  or  -f . 

2.  Given   \      xy  '    >   to  find  the  values  of  x  and  y. 

t  9y-9a;=:18,  ) 

(  y=4,  or  f . 
3«  Given  1    «  ,    '    ^/  *  *      '    '  {•  to  find  the  values  of  x  and  y, 

ly=4,  or  —  V- 

4i  Given  J    "~     ~         i    r  to  find  the  values  of  x  and  y. 
\  4—x  =y~y% ) 

,        Ans,   \  "=^  "^  ^ 
(2/=:l,  or|. 

5.  Given  i      i^^~[  ^'     f  to  find  the  values  of  a?  and  y. 

(  Sarfr— yi=14,  ) 

^71*     i  ^=(14)',  or  8, 

6.  Given  j    i  ,  ^Z       f  to  find  tlie  values  of  x  and  y. 

(  y=8,  or  0. 

7.  Giren  ^+J:+''2'=2y^+*'^' La^dthevaluesof  .andy. 

(     Vx+Vy=5,  ) 

(  y=4,  or  ^jS-. 

8.  Given   \y^     y  ~  9  ' )-   to  find  the  values  of  x  and  y. 

(  a;—  y  =2,    ) 


AFFECTED   QUADRATICS. 


299 


9«  Given  ^ 


Vx 


^x 


Vy 


>  to  find  X  and  y, 
x-\,ox\. 


Ans. 


(y=4,orJ/. 

o     .       ^  '    .,  }•  to  find  the  values  of  a;  and  y. 
y''  +  4x=2y-\-llS 

(aj=— 46,  or2, 
(  y=15,  or  3. 

11,  Given  j  h  aZ^qoo  f  *^  ^^^  *^®  values  of  a?  and  y. 

Ans     J  ^=35,  or -if^iL, 

(y=:16,orif4A 

'2a;  +  7y     ^       61+2a;  ^ 
=  2y— — T:r— , 

►  to  find  the  values  of  a:  and  y, 
i  y=4:,  or  Iff, 


12.  Given  ^ 


4a; 
4:X  +  Sy 


=y-2, 


13.  Given  -( 


16 
4:Xy+Sy-S 


4y-^3x—2     18— a; 


5a:  5 

3x-{-y_Sx—5y 
T~~~"3 


to  find  a;  and 

y- 


Ans     i^=^'^^-2l6, 
^^^-    1y=3,or|Hf 

14.  Given  j  ^  +  4V^^+4y=21  +  8f^y  +  4i/i^  j   to  find  ^  and  y. 
(  Vx-{yy=6,  )  \ 


♦  >'  »■ 


327.    QUESTIONS  PRODUCING  AFFECTED  QUADRATICS. 

QU  ESTIO  N. 

A  and  B  sold  130  yards  of  calico  (of  which  40  yards  were -4*8, 
and  90  ^'s),  for  $42.  Now,  A  sold  for  $1,  i  of  a  yard  more  than  B 
did.     How  many  yards  did  each  sell  for  $1  ? 


800  AFFECTED  QUADRATICS. 

Let  X  =  the  No.  of  yards  B  sold  for  $1. 
.«.  aj+»  =         «         «         ^       «        " 

-  =  what  B  got  for  1  yard. 


'    '     3 

90 

X 

40 


ii 


u 


u 


A      "        " 

B      "      90  yards. 


x+- 


^      «      40     " 


...     90  ,     40 

42=—  +  - 


X     x-\-\ 

^.     45  ,     20" 
21= h- 


21a;'  +  7ic=45a;  +  15  +  20aj 
21a;'— 58cc=l5 

a;= — ~ — =3,  or——  No.  of  yards  B  sold  for  |1. 
a;+i=3i  "  "      A       "        " 


QUESTIONS. 

1.  A  merchant  sold  a  quantity  of  cloth  for  |39,  and  gained  as 
much  per  cent,  as  the  cloth  cost  him.  What  was  the  price  of  the 
cloth?  Ans.  $30. 

2.  There  are  two  numbers  whose  difference  is  9,  and  their  sura 
multiplied  by  the  greater  produces  266.     What  are  the  numbers  ? 

Ans.  14  and  5,  or— 9i  and  —  18i. 

3.  It  is  required  to  find  two  numbers,  the  first  of  which  may  be  to 
the  second  as  the  second  is  to  16,  ;  and  the  sum  of  the  squares  of  the 
numbers  may  be  equal  to  225.  Ans.  9  and  12. 

4.  Bought  two  sorts  of  linen  for  6  crowns.  An  ell  of  the  finer  cost 
as  many  shillings  as  there  were  ells  of  the  finer.  Also,  28  ells  of  the 
coarser  (which  was  the  whole  quantity)  were  at  such  a  price  that  8 
ells  cost  as  many  shillings  as  1  ell  of  the  finer.  How  many  ells  were 
there  of  the  finer,  and  what  was  the  value  of  each  piece  ? 

Ans.  4  ells  of  the  finer ;  the  value  of  the  finer  16  shillings,  and 
of  the  coarser  14  shillings. 


AFFECTED  QUADRATICS.  .301 

6.  Two  partners,  A  and  B,  gained  |18  by  trade,  ^'s  money  was 
in  trade  12  months,  and  he  received  for  his  principal  and  gain  $26. 
J5's  money,  which  was  |30,  was  in  trade  16  months.  How  much  did 
A  put  in  trade  ?  Ans.  |20. 

6.  A  person  bought  some  sheep  for  $72,  and  found  that  if  he  had 
bought  6  more  for  the  same  money,  he  would  have  paid  $1  less  for 
each.     How  many  did  he  buy,  and  what  was  the  price  of  each  ? 

Ans.  18  sheep,  at  $4  a  piece. 

I.  The  plate  of  a  looking-glass  is  12  inches  by  18,  and  is  to  be 
framed  with  a  frame  of  equal  width,  whose  area  is  to  be  equal  to  that 
of  the  glass.     What  is  the  width  of  the  frame  ?         Ans.  3  inches. 

8.  There  are  two  square  buildings,  that  are  paved  with  stones,  a 
foot  square  each.  The  side  of  one  building  exceeds  that  of  the  other 
by  12  feet,  and  both  their  pavements  taken  together  contain  2120 
stones.     What  are  the  lengths  of  them  separately  ? 

Ans.  26,  and  38  feet,  respectively. 

9.  A  laborer  dug  two  trenches,  one  of  which  was  6  yards  longer 
than  the  other,  for  £l1  16s.,  and  the  digging  of  each  of  them  cost  as 
many  shillings  a  yard  as  there  were  yards  in  length.  What  was  the 
length  of  each  ?  Ans.  10,  and  16  yards. 

10.  A  company  at  a  tavern  had  £8  15s.  to  pay ;  but  before  the 
bill  was  paid,  2  of  them  sneaked  off,  in  consequence  of  which  those 
that  remained  had  each  10  shillings  more  to  pay.  How  many  were 
in  the  company  at  first  ?  Ans.  7. 

II.  A  grazier  bought  as  many  sheep  as  cost  him  £60;  out  of 
which  he  reserved  15,  and  sold  the  remainder  for  £54,  gaining  2 
shillings  a  head  on  those  he  sold.  How  many  sheep  did  he  buy,  and 
what  was  the  price  of  each  ?     Ans.  75  sheep,  at  16  shillings  each. 

12.  What  two  numbers  are  those  whose  sura  is  19,  and  whose  dif- 
ference multiplied  by  the  greater  is  60  ?  Ans.  12  and  7. 

13.  If  the  square  of  a  certain  number  be  taken  from  40,  and  the 
square  root  of  this  difference  be  increased  by  10,  and  the  sum  multi- 
plied by  2,  and  the  product  divided  by  the  number  itself  the  quotient 
will  be  4.     What  is  the  number  ?  Ans.  6. 

14.  A  person  being  asked  his  age,  answered,  if  you  add  the  square 
root  of  it  to  ^  of  it,  and  subtract  12,  there  will  remain  nothing.  How 
old  was  he?  Ans.  16,  or  36.     (Prove the  last  answer.) 

16.  A  and  JB  set  out  fr<^m  two  towns  which  were  at  the  distance  of 


302  AFFECTED  QUADRATICS. 

247  miles,  and  traveled  the  direct  road  till  they  met.  A  went  9  miles 
a  day ;  and  the  number  of  days,  at  the  end  of  which  they  met,  was 
greater  by  3  than  the  number  of  miles  which  B  went  in  a  day.     How 

many  miles  did  each  go?  Am.  ^117,  and  B  130. 

« 

16.-4  set  out  from  C  toward  i>,  and  traveled  7  miles  a  day.  After 
he  had  gone  32  miles,  B  set  out  from  JD  toward  (7,  and  went  every 
day  -jJg  of  the  whole  journey ;  and  after  he  had  traveled  as  many  days 
as  he  went  miles  in  one  day,  he  met  A.  What  is  the  distance  from  C 
toD?  Ans.  152,  or  76. 

17.  Three  merchants,  A^  B,  and  C,  made  a  joint  stock,  by  which 
they  gained  a  sum  less  than  that  by  $80.  ^'s  share  of  the  gain  was 
$60  ;  and  his  contribution  to  the  stock  was  $17  more  than  ^'s.  Also, 
B  and  C  together  contributed  $325.  How  much  did  each  contrib- 
ute? Ans.  A  $75,  ^$58,  (7  $267. 

18.  The  joint  stock  of  two  partners,  A  and  J5,  was  $416.  ^'s 
money  was  in  trade  9  months,  and  ^'s  6  months ;  when  they  shared 
stock  and  gain,  A  received  $228,  and  5  $252.  How  much  was  each 
man's  stock  ?  Ans.  A's  $192,  and  i5's  $224. 

19.  A  body  of  men  were  formed  into  a  hollow  square  3  deep, 
when  it  was  observed,  that  with  an  addition  of  25  to  their  number, 
a  solid  square  might  be  formed,  of  which  the  number  of  men  in  each 
side  would  be  greater  by  22  than  the  square  root  of  the  number  of 
men  in  each  side  of  the  hollow  square.  What  was  the  number  of  men 
in  the  hollow  square  ?  Ans.'  936. 

20.  A  merchant  bought  a  number  of  pieces  of  two  different  kinds 
of  silk  for  £92  3s.  There  were  as  many  pieces  bought  of  each  kind, 
and  as  many  shillings  paid  per  yard  for  them,  as  a  piece  of  that  kind 
contained  yards.  Now  2  pieces,  one  of  each  kind,  together  measured 
19  yards.     How  many  yards  were  there  in  each  ?     Ans.  11  and  8. 

21.  A  vintner  sold  7  dozen  sherry  and  12  claret,  for  £50.  He 
sold  3  dozen  more  of  sherry  for  £10  than  he  did  of  claret  for  £6. 
What  was  the  price  of  each  per  dozen  ? 

Ans.  Sherry  £2,  and  claret  £3. 

22.  ^  and  B  hired  a  pasture,  into  which  A  put  4  horses,  and  B  as 
many  as  cost  him  18  shillings  a  week.  Afterward  B  put  in  2  addi- 
tional horses,  and  found  that  he  must  pay  20  shillings  a  week.  At 
what  rate  was  the  pasture  hired  ?  Ans.  30  shiUings  per  week. 

23.  A  merchant  bought  54  gallons  Cognac  brandy,  a,nd  ^  certain 


AFFECTED  QUADRATICS.  803 

qtrantity  of  British.     For  the  former  he  gave  ^  as  many  shillings  per 

gallon  as  there  were  gallons  of  British,  and  for  the  latter  4  shillings 

per  gallon  less.    He  sold  the  mixture  at  10  shillings  per  gallon,  and  ^ 

lost  £28  16s.  by  his  bargain.     What  was  the  price  of  the  Cognac, 

and  the  number  of  gallons  of  British  ? 

Ans.  Cognac  18s.  per  gallon,  and  36  gallons  of  British. 
( 

24.  What  number  is  that,  which  being  divided  by  the  product  of 
its  two  digits,  the  quotient  is  2,  and  if  2*7  be  added  to  it,  the  digits 
will  be  inverted  ?  Ans»  36. 

25.  I  have  a  certain  number  in  my  mind  ;  this  I  multiply  by  2i, 
add  7  to  the  product,  and  multiply  this  sum  by  8  times  the  number ; 
I  now  divide  by  14,  and  subtract  from  the  quotient  4  times  the  num- 
ber, and  obtain  2352.     What  number  is  it  ?  Ans.  42. 

26.  What  two  numbers  are  those  whose  sum  is  100,  and  the  sum 
of  whose  square  roots  is  14  ?  Ans,  64  and  36. 

27.  What  niunber  is  that  which  if  you  subtract  from  10  and  mul- 
tiply the  remainder  by  the  number  itself,  the  product  shall  be  21  ? 

.   Ans.  7,  or  3. 

28.  What  are  the  two  parts  of  24  whose  product  is  equal  to  36 
times  their  difference?  Ans.  10  and  14. 

29.  What  two  numbers  are  those  whose  sum  is  8,  and  the  sum  of 
whose  cubes  is  152  ?  Ans.  3  and  6. 

30.  Two  partners,  A  and  B,  gained  $140  by  trade ;  ^'s  money 
was  3  months  in  trade,  and  his  gain  was  $60  less  than  his  stock,  and 
^'s  money,  which  was  $50  more  than  ^'s,  was  in  trade  5  months. 
What  was  ^'s  stock  ?  Ans.  $100. 

31.  What  two  numbers  are  those,  the  difference  of  whose  squares 
is  q^,  and  which  being  multiplied,  respectively,  by  a  and  6,  the  differ- 
ence of  the  products  is  s^  ? 

,        as'±:bVs*-(a'-b')q^ 
Ans.  r — )-^ ^^. 


,  bs''dtzaVs'-ia^-b'')q'' 

and  5 — ^^ '—. 

a  —b 


32.  Into  what  two  parts  can  a  and  b  each  be  divided,  such  that  the 


304  AFFECTED  QUADRATICS. 

product  of  one  part  of  a  by  one  part  of  b  shall  be  p,  and  the  product 
of  the  remaining  parts  ^  ? 

^li_ab-(a-p)±:Vjab~-(g-p)i'-4ahp 


Ans,    -( 


^ab  +  {^-p)^)/\ab-{g-p)Y-iabp 


ii^rjsr^z^ 


_ab-(a    p)dzV\ab-(A-p)Y-4aAp 
^ab-\-{h-p)±:V{ab-{§-^'-4.abp 

33.  During  the  time  that  the  shadow  of  a  sundial,  which  shows 
true  time,  moves  from  one  o'clock  to  five,  a  clock  which  is  too  fast  a 
certain  number  of  hours  and  minutes,  strikes  a  number  of  strokes  = 
that  number  of  hours  and  minutes,  and  it  is  observed  that  the  number 
of  minutes  is  less  by  41  than  the  square  of  the  number  which  the 
clock  strikes  at  the  last  time  of  striking.  The  clock  does  not  strike 
twelve  during  the  time.    How  much  is  it  too  fast  ? 

Ans,  3  hours  and  23  minutes. 

34.  A  and  B  engage  to  reap  a  field  for  £4  10s. ;  and  as  A^  alone, 
could  reap  it  in  9  days,  they  promise  to  complete  it  in  5  days.  They 
found,  however,  that  they  were  obliged  to  call  in  (7,  an  inferior  work- 
man, to  assist  them  for  the  2  last  days,  in  consequence  of  which  B 
received  3^.  9c?.  less  than  he  otherwise  would  have  done.  In  what 
time  could  Bj  or  (7,  alone,  reap  the  field  ? 

Ans.  B  in  15,  and  (7  in  18  days. 

35.  There  are  three  numbers,  the  difference  of  whose  differences  is 
8 ;  their  sum  is  41 ;  and  the  sum  of  their  squares  699.  Wliat  are 
the  numbers?  Ans.  7,  11,  and  23. 

36.  There  are  three  numbers,  the  difference  of  whose  differences  is 
5  ;  their  sum  is  44  ;  and  their  continued  product  is  1950.  What  are 
the  numbers?  Ans.  6,  13,  and  26. 

3*7.  A  grocer  sold  80  pounds  of  mace,  and  100  pounds  of  cloves  for 
$65 ;  but  he  sold  60  pounds  more  of  cloves  for  $20  than  he  did  of 
mace  for  $10.     What  was  the  price  of  a  pound  of  each  ? 

Ans.  Mace  50,  and  cloves  25  cents  a  pound. 

38.  The  fore-wheel  of  a  carriage  makes  6  revolution  more  than  the 
hind- wheel  in  going  120  yards;  but  if  the  periphery  of  each  wheel 


AFFECTED   QUADRATICS.  805 

be  increased  one  yard,  it  will  make  only  4  revolutions  more  than  the 
hind-wheel  in  the  same  space.    What  is  the  circumference  of  each  ? 

Ans.  4  and  5  yards. 

39.  ^  and  B  were  going  to  market,  the  first  with  cucumbers,  and 
the  second  with  3  times  as  many  eggs ;  and  they  find  that  if  B  gives 
all  his  eggs  for  the  cucumbers,  A  would  lose  10  pence,  according  to 
the  rate  at  which  they  were  selling.  A,  therefore,  reserves  |  of  his 
cucumbers,  by  which  B  would  lose  sixpence,  according  to  the  same 
rate.  But  B^  selling  the  cucumbers  at  sixpence  apiece,  gains  upon 
the  whole  the  price  of  6  eggs.  What  was  the  number  of  eggs,  and 
cucumbers,  and  the  price  of  one  of  each  ? 

'~Ans.  30  eggs  at  1  penny  a  piece,  and  10  cucumbers  at  4  pence 
_  apiece. 

40.  A  person  bought  a  certain  number  of  larks  and  sparrows  for 
6  shillings.  He  gave  as  many  pence  per  dozen  for  larks  as  there 
were  sparrows,  and  as  many  pence  per  score  for  sparrows  as  there 
were  larks.  If  he  had  bought  10  more  of  each  (the  price  of  the  larks 
remaining  the  same),  and  had  given  as  much  per  dozen  for  sparrows 
as  he  gave  per  score  for  larks,  they  would  have  cost  £l  5s.  5d.  What 
was  the  number  of  each  ?  Ans,  15  larks,  and  36  sparrows. 

41.  A  poulterer  bought  a  certain  number  of  ducks  and  18  turkeys 
for  $110 :  each  turkey  costing  within  1  dollar  as  much  as  three  ducks. 
He  afterwards  bought  as  many  ducks  and  5  more,  and  20  turkeys, 
giving  one  dollar  a  piece  more  for  each  duck  and  turkey  than  before ; 
and  found  that  the  value  of  his  former  purchase  was  to  the  value  of 
the  latter  one  : :  2:3.  What  was  the  number  of  ducks,  and  the  prices 
of  the  ducks  and  turkeys  at  the  first  purchase  ? 

Ans.  10  ducks,  and  the  price  of  a  duck  $2,  and  of  a  turkey  |5. 

42.  A  and  B  put  out  difierent  sums  at  interest,  amounting  together 
to  $200.  B^s  rate  of  interest  was  1  per  cent,  more  than  ^'s.  At 
the  end  of  5  years,  B^s  accumulated  simple  interest  wanted  but  $4  to 
be  double  of  ^*s.  At  the  end  of  10  years,  A^&  principal  and  interest 
was-to  ^'s  as  5:8.  What  was  the  sums  put  out  by  each,  and  the 
rate  per  cent.  ? 

Ans.  A  put  out  $80  at  5  per  cent.,  and  B  $120  at  6  per  cent. 

43.  When  the  price  of  brandy  was  3  times  the  price  of  British 
spirit,  a  merchant  made  two  mixtures  of  brandy  and  British  spirit, 
and  the  prices  per  gallon  were  in  the  ratio  of  9  to  10.  He  afterwards 
mixed  twice  as  much  brandy  with  the  same  quantity  of  British  spirit 

20 


806  AFFECTED  QUADRATICS. 

in  each  case,  and  the  relative  prices. were  the  same  as  before.    "What 
was  the  ratio  of  the  quantities  mixed  ? 

Ans.  The  first  mixtm*es  were  as  3  to  1,  and  2  to  1,  and  the  second 
as  3  to  2,  and  1  to  1. 

44.  A  and  B  traveled  on  the  same  road  and  at  the  same  rate  from 
Huntingdon  to  London.  At  the  50th  milestone  from  London,  A  over- 
took a  drove  of  geese  which  were  proceeding  at  the  rate  of  3  miles 
in  2  hours,  and  2  hours  afterwards  met  a  stage-wagon,  which  was 
moving  at  the  rate  of  9  miles  in  4  hours.  B  overtook  the  same  drove 
of  geese  at  the  45th  milestone,  and  met  the  same  stage-wagon  ex- 
actly 40  minutes  before  he  came  to  the  31st  milestone.  Where  was 
B  when  A  reached  London  ?  Am.  25  miles  from  London, 

45.  The  difference  between  the  hypothenuse  and  base  of  a  right- 
angled  triangle  is  6,  and  the  difference  between  the  hypothenuse  and 
the  perpendicular  is  3.     What  are  the  sides?     Ans.  15,  12,  and  9. 

46.  In  digging  among  some  ruins,  the  workmen  found  9  urns, 
together  containing  60  gold  coins ;  the  2d  and  8th  containing  8  and 
4  respectively.  They  secreted  a  certain  number  of  these,  greater 
than  the  number  they  left ;  which  being  afterwards  recovered,  it  was 
found  that  the  number  of  urns  secreted  was  to  the  number  left  as  the 
number  of  coins  secreted  was  to  the  number  remaining.  Now  if  in- 
stead of  taking  the  2d  urn,  they  had  carried  off  the  8th,  then  the 
number  of  coins  taken  away  would  have  been  to  the  number  remain- 
ing as  the  square  of  the  number  of  urns  secreted  to  the  difference 
between  that  square  and  20  times  the  number  of  urns  remaining. 
What  number  of  each  was  secreted  ? 

A71S.  6  urns  and  40  coins. 

47.  Bacchus  caught  Silenus  asleep  by  the  side  of  a  full  cask,  and 
seized  the  opportunity  of  drinking,  which  he  continued  for  |  of  the 
time  that  Silenus  would  have  taken  to  empty  the  whole  cask. ,  After 
that  Silenus  awoke,  and  drank  what  Bacchus  had  left.  Had  they 
drank  both  together,  it  would  have  been  emptied  2  hours  sooner,  and 
Bacchus  would  have  drank  only  ^  what  he  left  Silenus.  How  long 
would  it  have  taken  each  to  have  emptied  the  cask  separately  ? 

Ans.  Silenus  in  3  hours,  and  Bacchus  in  6. 


CHAPTER  XII.  / 

CUBIC  EQUATIONS. 

(328«)  Cubic  Equations  may  be  divided  into 
Pure  Cuhics^ 
Incomplete  Cuhics,  and 
Affected  Cubics, 
(3 3 9.)  A  pure  cubic  equation  is  one  in  which  the  unknown 
quantity  is  contained  in  but  one  term,  and  whose  exponent  is  3,  or  a 
fraction  which  when  reduced  to  its  lowest  terms  the  numerator  is  3 ;  as, 

3.  S. 

ax^=dbb;  ax^=:±:b*  ax^=^b,&G, 
(330*)  The  solution  of  a  pure  cubic  presents  no  difficulty. 

PROBLEM. 

Given  3a;' =24  to  find  x. 

SOLUTION. 

3ic''=24, 

a;''— 8=0, 
(a;~2)(a;'  +  2a:4-4)=0. 
Whence  we  get  ar — 2 = 0, 
or  ic"  4- 2a; +  4=0, 
whose  solutions  give  a;=2,  or  —  Izfc  1^— 3. 
If  but  one  value  of  the  unknown  quantity  were  required,  we  might 
in  the  equation  a;' =8  merely  take  the  cube  root  of  both  sides,  and 
obtain  a:=2. 


EXAMPLES. 

1.  Given  a;' =  27  to  find  x. 

Ans,  a:=3,  or . 

2 

2.  Given  ia;''=32  to  find  x. 

Ans,  a;=4,  or  — 2±2|/— 3. 

3*  Given  a;' =4  to  find  x. 

Ans. 

a;=V4,  or  -iV4±|/~fV2. 

308  CUBIC  EQUATIONS. 

4.  Given   )    ^    K'^Z^"    '^\  to  find  the  values  of  a;  and  y. 

{  X  — y  =00  ) 

Ans,    i    ~  '  ,     , . 

(y=2,  or  — l±f/—3 


5. 


!•  Given  -j      ,    '    '/     '     \  to  find  the  values  of  a?  and  v. 
( iry''=384  )  ^ 

,     U=24, 


^715.    '-=24,or-12±12i/-3 


or  — 2db2K— 3 

(331.)  An  incomplete  cubic  equation  is  one  in  which  the  uaiknown 
quantity  is  contained  in  but  two  terms :  the  exponent  of  one  being  3, 
and  the  other  1  or  2  ;  or,  the  exponent  of  one  being  3  times  a  proper 
fipaction  whose  numerator  is  1,  and  which  is  either  the  exponent  or 
half  the  exponent  of  the  other  term ;  as, 
ax^-±.hx=iizc ;  ax^:hbx^=±:c;  ax±:bxi=dcc ;  ax±bx^=d:c ;  SsG. 

Remaek. — 1^0  general  practicable  method  of  obtaining  the  values  of  the  un- 
known quantity  in  incomplete  or  affected  cubics  has  as  yet  been  discovered,  and 
we  are,  therefore,  compelled  to  resort  to  what  are  called  specicU,  tentativef  or 
approximative  methods. 

Some  of  these  methods  are  here  set  forth. 

PROBLEM. 

Given  a;'-i-3a;=14  to  find  the  values  of  w, 

SOLUTION. 

a:'  +  3ar=14  (1). 

X*  +  3a;''  =  1 4a;  (2) = (1)  x  a;.  [members. 

a;*  +  '7a;''=4a;''  +  14a;       ^  (3)=(2)  with  4a;'' added  to  boHi 

a.*4.'ra;''  +  (l)'=4a;''+14a;  +  (f)'.  (4)=(3)  square  completed, 

a;"  +  f = 2a;  + 1,  or  -2a;-|,      (5) = Vjl). 

.'.     a;=:2  ;         or  a;''4-2a;=  — *?      

a;=— 1±1/— 6. 

EXAMPLES. 

1.  Given  x^—^x=-~Q  to  find  x,  Ans.  a;=l,  2,  or  —3. 

2.  Given  a;'— 32a;=:  —24  to  find  x.        Ans.  x=z—Q^  or  3 ±4/6. 

3.  Given  a;'— 22a;=24  to  find  x.  Ans.  a;=— 4,  or  2±V^10, 

4.  Given  a;'  +  6a;=:88  to  find  x.       Ans.  a;=4,  or  —2dtsV'^. 

5»  Given  a;'+6a;=45  to  find  x.       Ans.  x=3,  or . 


CUBIC  EQUATIONS.  309 

6*  Given  ic'— 13a;=— 12  to  find  x.  Ans.  x=l,  3,  or  —4. 


7.  Givena;'  +  48a;=:104  to  find  x.    Ans,  a;=2,or  —ldzV—51. 

8.  Given  x'—6x=9  to  find  x.  Ans,  x=3^  or . 


9.  Given  a;  +  Tari-=22  to  find  x,  Ans,  a:=8,  or  29rh'7'/— 10. 

10.  Given  a;'+3a;=14  to  find  x,  Ans,  x—2,  or  —iztV^, 

11.  Given  x^ =14  to  find  x.  Ans.  x= — ,  or  . 

3x       ^  3'            8 

12.  Given  a;"— 2a;=:  — 4  to  find  x,  Ans,  a;=:— 2,  or  Idbt^  — 1. 


13.  Given  5a;''4-2a;=44  to  find  x,      Ans.  x=2,  or — . 

6 

14.  Given  a;'+108a;=665  to  find  x.    Ans,  x=5,  or^^^^^~^ 

2 

15.  Given  x^—S9x=  —  10  to  find  x,  Ans,  x=2,  5,  or  —7. 

16.  Given  a;"— 49a;=120  to  find  x,  Ans.  x=S,  —3,  or  —5. 

17.  Given  a;'+12ar=-63  to  find  x,    Ans,  a;=— 3,  or^'^^^~^ 

2 

18.  Given  a;''— 21a;=— 344  to  find  x, 

Ans.  a;=— 8,  or  4±3f"^. 

19.  Given  a;'— 6a;=40  to  find  x.  Ans,  a;=4,  or  —2=1:^—6. 

20.  Given  x'-1^x=-^2Q0^  to  find  x, 

Ans,  .=^7,or-^^ 
'  2 

(332.)  Sometimes  the  factors  of  an  equation  are  very  apparent, 
and  then  the  root  may  be  obtained  as  in  the  solution  to  the  following 

PROBLEM. 

Given  a;"— 3a;— 2=0  to  find  the  value  of  a?. 

SOUJTION. 

«'— 3a;— 2=0 

x^~x=2x+2 
a;(a;'-l)=2(a;4-l) 


810  CUBIC  EQUATIOKS. 

Since  a;  4- 1  is  a  factor  of  both  members  of  the  equation,  the  equatioD 
will  be  satisfied  by  putting       a;+ 1  =0 

Whence,         x=  —  l. 
To  get  the  other  values  of  x  divide  both  members  by  a;+l. 
And  then,     x^—x=2 

x=2,  or  — 1. 

KXAMPtES.  

I,  Given  a;'— 5a;=— 4  to  find  x.         Ans,  a;=l,  or 


2.  Given  a;'— 2a;=l  to  find  x,  Ans.  a;=  — 1,  or 

3*  Given  re"— 2a;=  — 1  to  find  x.  Ans.  x=l,  or  — 


2 

2      ' 

l±V/5 


2 

4*  Given  ic^— 8aj=  — 8  to  find  x.  Ans,  x=2,  or  —  Irhi^S. 

2 

5*  Given  a;--l=2H — j-  to  find  a?.  -4/is.  a;=l,  1,  or  4. 

arg- 

x'  -{-Sx 7 

6t  Given  7^=1  ^  ^^  ^'  ^^'  ^=3,  —2,  or  —3. 

a;  +  2+  — 
a; 

7.  Given  3x^—^x^=1x^3  to  find  a:.  Ans.  x=^,  3,  or  —1. 

8.  Givena;'  +  a;'  +  a;  +  l  =  10a;  +  10  tofinda;. 

Ans.  x=S,  —1,  —3. 

9.  Given  a;'— a;' --2a;  =—2  to  find  x.  Ans.  a?=l,  or  ±4^2. 

10.  Given  (a:''4-2ar)(a;+4)=2— (a;  +  4)  to  find  x. 

Ans.  x=—2y  or  — 2±i^3. 

i3x-—=v'-v  ) 

11,  Given   •<  y     ^      ^'  S   to  find  a;  and  y. 

(     y+a;=4.         ) 

.        j  a;=3,  or  1. 

(333*)  It  sometimes  happens  that  the  coefficients  of  the  unknown 
quantity  have  such  relations  that  the  equation  may  be  reduced  ac- 
cording to  the  method  adopted  in  the  solution  of  the  following 

PROBLEM. 

Givena;'— 6aj'  +  lla;=r6  to  find  the  values  of  x. 


CUBIC  EQUATIONS.  311 


SOLUTION. 

aj'-6a;''4-lla;-6=0 

x'-6x'  +  Ux'-ex=0 

x'—6x'-^9x'+2x''—6x=0 

(x'-'SxY-{-2(x'-Sx)=0 

(a) 

(a5*-3a;)'  +  2(a5'-3a;)  +  l=l 

x''-Bx+l  =  ±l 

x^—Sx=0,  or  ■ 

-2, 

.-.    x'=Sx; 

or  aj^— 3aj= 

2 

05=3  05=— ^r— =2,orl. 

In  equation  (a)  put  y=zx^—Sx,  and  we  shall  have 
y'»  +  2y=0 
y'»  +  2y  +  l=l 

2/=0,  or  -2, 
•.     a;'— 3£c=0  ;  or  —2,  the  same  as  before. 
Remark. — Judicious  substitution  often  saves  figures,  and  also  often  reveala 
relations  which  might  not  otherwise  he  so  apparent 

Since  a:"— 3a?  is  a  factor  of  (a),  we  have  immediately 

a;'— 3a;=0,  and  by  dividing  by  this, 
a;"— 3a:+2=0,  the  same  as  before. 

EXAMPLES. 

1.  Given  x''—6x'  4-  6a;  + 12=0  to  find  x.     Ans,  a;=  —  1,  3,  or  4. 

2.  Given  ar^— 2a;''— a;  +  2-=0  to  find  x.        Am.  x=l,  2,  or  —1. 

3.  Given  aj'— Ga^'  -f  12a:— 9=0  to  find  x.  

.             „       SdtV-3 
Ans.  a;=3,  or , 

4.  Given  a;'  — 2a;''— 5a; +  6=0  to  find  ar.  ^w«.  a;=l,  3,or  — 2. 
5«  Given  x^  +  2a;''— 6a;— 6=0  to  find  x,  Ans.  a;=2,  —1,  or  —3. 
6*  Given  a;''  +  6a;''  +  lla;-!-6=0  to  find  x. 

Ans.  a;=  — 1,  —2,  or  —8. 
7t  Given  a;'— 8a;''  +  19a;— 12=0  to  find  x.  Ans.  a;=l,  3,  or  4. 
8.  Given  a;'  +  2aa;''  +  5a''a;+4a' =0  to  find  x.  

—a±aV— 15 

Ans.  x=—aj  or . 

2 

(334.)  The  following  examples  may  be  solved  according  to  the 

artifices  employed  in  quadratics. 


812  CUBIC  EQUATIONS. 

PROBLEM . 

Given  a;'— 3a;' +  3a; =9  to  find  the  values  of  x. 


\ 


SOLUTION. 

a;'— 3a;*4-3a;=9         (1). 
a.«_3a;a  4-  Sx— 1=8         (2) =(1)  with  —  1  added  to  both  members. 
a;-l=2         (3)=V(2). 
a;=3 
To  obtain  the  other  values  find  the  other  factor  of  a;'— 3aj''4-3a;— 9, 
SB— 3  being  one.    That  factor  is  ar*  +  3  ;  hence,  we  have 

a;'' +  3=0. 

3;^^  =— 3. 

a;=±4/^. 
The  best  method  of  solving  this  equation  is  by  factoring,  for  its 
factors  are  very  apparent. 

a;'— 3a;''  +  3a;— 9=0. 

a;Xa;— 3)+3(a;— 3)=0. 

whence,  (a;''+3)(a;-3)=0. 

a;— 3=0. 

and  a;' +  3=0. 

whence,       a;=3,  or  ±4/— 3. 

Remaek. — ^The  examples  which  follow  are  not  all  of  the  character  of  the  one 
whose  solution  has  just  been  given,  but  embody  different  principles  with  which 
the  student  is  supposed  to  be  familiar. 

EXAMPLES. 

1,  Given  a;'  +  6a;'+12a;=19  to  find  x. 


.            ,         -7  +  3*^3 
Ans.  x=lj  or . 


2,  Given  >|/a;''  +  8=*/125— 6a;'  — 12a;  tofinda;. 


J             Q          —9±  51^-3 
Ans.  a;=3,  or . 


3.  Given  Va;"— a'=N/3aa;'— 3a»a;  +  86tofinda;. 

Ans,  a;=a+2V6. 

J    ^.       Va  +  x    Va+x    Vx  ^    ^   ,  .   ■  acl 

4i  Given 1 = — to  find  x,  Ans.  x: 


5.  Given  a;'— 6a;' + 12a; =8  to  find  x.  Ans.  a;=2,  2,  2. 

6*  Given  a;^— 16a;' +  76a;= 12 6  to  find  x.  Ans.  x=z5j  5,  6. 


CUBIC   EQUATIONS.  BIB 

HE NKLE'S  METHOD.* 
(335.)  The  following  tentative  process  for  solving  affected  and  in- 
jomplete  cubics,  has  never  before  been  given  in  a  work  upon  algebra. 

PROBLEM. 

1.  Given  a;"— 8«''  +  lla;  +  20=0  to  find  the  values  of  a;. 


SOLUTION. 

a;3_8a;3^]l:c_^20  =  0. 

a;*_8a;'  +  lla;''  +  20a;=0. 

{x^—4xy—5x^  +  20x=0. 

{A), 

(x''-4:Xy-4:{x'-4:X)+4=X'-4:X-^i, 

(B). 

x^-'4x—2=:x—2,  or  2—x, 

x^=5x\  or  x^—Bx= 

=4. 

x=5                     x- 

=-— =4,01-1, 

This  solution  vsdll  need  some  explanation.  Every  affected  cubic 
equation  may  be  put  into  the  form  expressed  by  (A)  that  is,  made  to 
cjonsist  of  the  square  of  a  binomial  followed  by  two  terms.  If  these  two 
terms  contain  as  a  factor  the  first  power  of  the  binomial,  the  equation 
may  be  solved  as  in  (B) ;  but  if  not,  we  may  proceed  as  in  the  above 
example.     Let  us  resume  the  equation 

{x''—4xY—bx^  +  20a;=0. 

Let  us  see  whether  x"^  is  not  the  first  term  of  a  binomial  square, 
which  being  added-  to  both  members  of  this  equation  will  render 
them  perfect  squares.  If  x"^  is  added  to  both  members,  we  see  that 
the  coefficient  of  the  second  term  in  {B)  is  —  4 ;  considering  {x^—4xy 
as  the  first  term ;  and  this  coeflScient  shows  that  the  third  term  in  the 
first  member  of  (B)  must  be  4,  and  therefore,  the  third  term  in  the 
second  member  must  also  be  4  ;  but  if  the  first  and  the  last  term  of  a 
binomial  square  are  x^  and  4,  the  middle  term  must  be  either  +  4ar, 
or  —4x.  From  which  we  see  that  the  square  of  the  binomial  of 
which  x^  is  the  first  term  must,  in  this  case,  be  x'^zh4x  +  4.  Since,  4 
is  in  both  members,  it  follows  that  if  a;'' ± 4a:  be  added  to  —  5a;''  +  20a;, 
and  the  result  should  be  —  4(a;^— 4a;),  or  —4a;'' +  16^,  that  a;' ±  4a;  4- 4 
is  the  proper  square  to  add.  As  we  have  already  addei  a.-*,  it  only 
remains  for  us  to  see  whether  4a;  added  to  or  subtracted  from  +  20aJ 

*  This  method  of  solvinp^  Cubic  Equations,  which  is  also  apphed  to  the  solu- 
tion of  Biquadratics  was  discovered  by  Prof.  W.  D.  Henkle.  1,  therefore,  have 
assumed  the  respousibilitj  of  calling  it  the  "Henkle  Method." — J.  F.  S. 


314 


CUBIC   EQUATIONS. 


makes  +  16a;.  It  will  be  seen  that  4x  subtracted,  or  what  is  the  same 
—4x  added  makes  +20a;  become  -f  16  a;,  therefore,  a;'— 4a; +  4  is  the 
proper  square  to  add. 

The  student  in  passing  from  (A)  to  (B)  should  put  his  trial  work 
one  side. 


<!> 

4 

I— 1 

^ 

I— 1 

ex\<ft 

, 

Si 

*«— ' 

00 

I-H 

4- 

+ 

T 

+ 

T 

+ 

1 

J? 

s 

^ 

o 

^ 

^ 

54 

Q 

CO 

CO 

t^ 

i^ 

Tj< 

»— 1 

S 

-H 

■n 

-H 

-H 

•V4 

i 

^ 

05 

CO 
r— 1 

CO 
CO 

% 

c« 

g. 

CI 

CT 

"^ 

% 

.i 

I— 1 

3 

-«o 

w 

00 

r^ 

►AS 

+ 

+ 

T 

4- 

+ 

+ 

^<a^ 

,,— ^ 

/^"^ 

x<^ 

« 

'^ 

'IT 

?^ 

« 

54 

+ 

+ 

+ 

+ 

+ 

+ 

*! 

%i^ 

5t 

,  Jl. 

Jt. 

2i, 

5*^ 

5^ 

CO 

o 

lO 

00 

Oi 

<N 

CO 

00 

cq 

rH 

^ 

1 

1 

1 

1 

1 

+ 

o 

H 

-^ 

y^'>. 

1 

S- 

5i 

1 

O 

QD 

l-H 

J 

II 

1 
II 

d 

+ 

l-H 

1 

« 

S 

54 

1> 

1 

i-T 

I— 1 

+ 

o 

<o 

o 

CO 
00 

1 

CO 

1 

-H 

C4 

.g 

II 

II 

II 

II 

II 

II 

1 

00 

-^ 

« 

54 

1— 1 

r-t 

^ 

54 

1 

1— 1 

+ 

CO 
CO 

rH 

ft 

+ 

+ 

I 
-54 

1l 

+ 

+ 
CO 

+ 

+ 

+ 

« 

i 

CO 

CO 

^ 

. 

1 

1 

'• 

CJ 

+ 

+ 

2^ 

H^ 

The  work  upon  the  right  shows  the  trial  work.  First  x*  was  added, 
which  resulted  in  ±3  3a;  for  the  middle  term  for  the  second  member, 
but  since  +33a;  or  —33a;  added  to  +14a;,  does  not  give  —33a;,  we 
see  that  a;?  is  not  the  proper  term  to  add  in  the  start.     So  when  4a;' 


CUBIC  EQUATIONS.  815 

is  added,  +60a;  or  —60a;  added  to  +14a;  does  not  give  —30a;;  or 
when  9a;'  is  added,  -\-^5x,  or  —Ibx  added  to  +14a;  does  not  give 
—25a;;  or  when  16a;Ms  added,  +I2x^  or  —'72a;  added  to  4-14a;  does 
not  give  —18a;;  or  when  25a;''  is  added,  +  45a;,  or  —45a;  added  to 
+  14a;  does  not  give  —9x ;  but  when  36a;M8  added,  —12a;  added  to 
+  14a;  does  give  +  2a;. 

The  failure  in  adding  a;'  is  so  much  that  the  operator  would  not 
likely  try  the  successive  squares,  but  would  immediately  pass  to 
16a;»,  25a;»  or  36x\ 

Eemark. — The  above  is  given  to  show  the  mode  that  the  student  should  pur* 
sue  in  finding  the  proper  square.  One  who  is  skillful  in  this  mode  of  solution 
seldom  makes  more  than  three  trials,  and  frequently  but  one. 

If  the  first  term  of  the  equation  is  x^  and  the  coefficient  of  the  second  term  is 
odd,  multiply  each  term  by  4x. 

If  the  first  term  is  not  a  square,  it  should  be  made  one  by  multiplying  the 
equation  by  four  times  the  coefficient  of  the  first  term.  When  the  coefficient  of 
the  second  is  even,  it  is  frequently  only  necessary  to  multiply  by  once  the  coefficient 
of  the  first  term.  It  is  not  absolutely  necessary  to  follow  the  last  directions 
given,  but  if  they  are  followed,  the  student  will  be  saved  the  trouble  of  operat- 
ing with  fractions  which  he  might  otherwise  encounter. 

PROBLEM 

3.  Given  a;' +  6a?' +  3a;— 9=0  to  find  the  value  of  a?. 

SOLUTION. 

a;'+5a;»  4- 3a;— 9=0. 
4a;*  +  20a;'+12a;''-36a;=0. 
(2a;''+5a;)'  — 13a;— 36a;=0. 
(2ar'  +  5a;)'— 6(2a;'+5a;)  +  9=a;='  +  6a;  +  9. 

2a;''  +  5a;— 3=a;  +  3,  or  —a;— 3, 

a;'  +  2a;=3;  or2a;'=  — 6aJ, 

a;=  — 1±2  =  1,  or  — 3  a;=— 8. 

PROBLEM 

4.  Given  3a;''  —  14a;'-f  21a;— 10=0  to  find  the  value  of  x. 

SOLUTION. 

3a;»— 14a;'' +  21a;— 10=0. 

9a;*— 42a;'  +  63a;'— 30a;=0. 

(3a;'  — Va;)'' +  14a;''— 30a;=0. 

{Bx^--1xY  +  5(Sx'-'7x)  +  'i\'=:x'-5<c-{-^\\ 

3a;''  — 7a;+f =a;— f,  or  f — a?, 

8a;*— 8a;=— 5;  or  3a;' = 6a?, 

4±1 
a;=— -— =4  or  1  a;=2. 

3         ' 


B16  CUBIC  EQUATIONS* 

(336.)  Since  the  solving  of  cubics  by  tbis  process  is  a  good 
mental  exercise,  a  copious  list  of  examples  is  appended.  The  pupil 
before  commencing  should  make  himself  familiar  with  the  perfect 
squares  from  1  to  1296. 

EXAMPLES. 

1.  Givena;'— 6a;'+13a?— 10=0  tofinda;.  

Ans.  x=:2,  or  2±:V'-1, 

2.  Given  ar'  +  6a;'— 7a;— 60=0  tofind  x.    Am.  x=S,  —4,  or  —6. 

3.  Given  4a^— 24a:»  +  46a;-26=0  to  find  a;. 

Ans.  x=zl,  2^,  or  2^. 

4.  Givena;'— 9a;'— 130a;  +  600=0  tofinda;. 

Ans.  a;=4,  16,  or  —10; 

5.  Given  a;' +  6a;'— 32=0  to  find  x.        Ans.  a;=2,  —4,  or  —4. 

6.  Givena;'— llaj'4-43a;— 65=0  tofinda;. 

Ans.  x=5^  or  3±y— 4. 

7.  Given  a;"— 233;*  + 167a;— 386=0  to  find  x. 

Ans.  x=5,  7,  or  11. 

8.  Given  a;»— 12a;'-f-36a;— 7=0  to  find  x. 

5dbV2i 
Ans.  a;=7,  or  — - — . 
2 

9.  Givena;"  +  8a;'-f  I7a;+10=0tofinda;. 

Ans.  a;=  — 1,  —2,  or  —5. 

10.  Given  a;'4-6a;'+20a;  +  16=0  to  find  x.  

-6=bi/335 
Ans.  a;=  — 1,  or . 

11.  Given  24a;'— 26a;' +  9a;— 1=0  to  find  a;.     ^W5.  a;=|,  ^,  or  ^. 

12.  Given  a;'- 24a;' +  191a; -604=0  to  find  x. 

Ans.  a; =7,  8,  or  9. 

13.  Givena;'  — 18a;'  +  49a;— 46  =  0  tofinda;. 

Ans.  a;=5,  or  4=bf^7. 

14.  Given  2a;'— 8ar*  + 2a?— 3=0  to  find  a;.  

Ans.  a;=li  or  zHdl. 

15.  Given  2a;'— 16a;' +  26a; +  6=0  to  find  a;.  _ 

9±V9l 
Ans.  a;=3,  or . 


CUBIC    EQUATIONS.  317 

16.  Given2a;»—12a;'+13a;— 16=0  to  find  iP. 

Ans,  x=5,  or . 

'  2 

17.  Givenaj'  +  lOa?"— lOla?— 990=0tofinda?. 

Ans.  a;=10,  —9,  or  —11. 

18.  GiTenaj"— 2a;'— 83a;+90=0tofinda?. 

Ans.  a: =3,  6,  or  —6. 

19.  Given  4a;'— 24a;»  +  21a;— 5=0  to  find  x.     Ans.  x=i,  i,  or  5, 

20.  Given2a;'— 33a;''  +  121a;  +  84=0tofinda;. 

Ans,  a?=7,  or . 

21.  Given  a;»+a;'— 34a;+66=0  to  find  x.    Ans,  a;=2,  4,  or  —% 

22.  Given  a;'—21a;'4- 146a;— 336=0  to  find  a;. 

Ans,  a;=6,  7,  or  8. 

23.  Given  a;'— •3a;' 4- '004=0  to  find  x.     Ans.  a;=-2,  -2,  or  — -1. 

21.  Givena;"— 103;"  + 10a;— 100=0  to  find  a;. 

Ans.  x=l%  or  ±f/— 10, 

25.  Given  a;'-2a;»+3a;-4^=0  to  find  a;.  

Ans,  a;=lf,  or . 

26.  Givena;'-4a;''— 28a;— 32=0tofinda;. 

Ans,  x=S,  —2,  or  —2. 

27.  Given  2a;'— 15a;*  +  26a;— 6=0  to  find  x, 

.  „        lldb^97 

Ans,  a;=2,  or . 

4 

28.  Given  a;' -a;'-'ra;  + 16=0  to  find  a;. 

Ans,  x=—Sj  or  2 ±1^-1. 

29.  Given2a;'— 6a;' +  4a;— 120=0  to  find  a;. 

J[w5.  x=5,  or  — l±f^— 11. 

30.  Givena;'  +  9a;'— 82a;— Y20=0  tofinda;. 

Ans.  a;=9,  —8,  or  —10. 

31.  Given  a;'— a;'— 10a; +  6=0  to  find  x. 

Ans.  a;=  — 3,  or  2±V2. 

32.  GivenSa;'- 26a;'-flla;+10=0tofinda;. 

.  „,         3=fcf^41 

Ans.  a;=2^,  or . 

2'  8 


818  CUBIC   EQUATIONS. 

33.  Given  2a.'  -|-12a!*+48a;~Y90=0  to  find  a?. 


.              ^        -ll±i/-195 
Ans,  x=5f  or . 


34.  Given  2a;'— 3a:' + a?— 30=0  to  find  x. 


Ans,  a;=3,  or 

4 

35.  Givenaj'  +  lYa;'— 140a;— 2496=0  to  find  a;. 

Ans,  a;=12,  —13,  or  —16. 

36.  Given  a;'4-3a;'— 64=0  to  find  x,    Ans,  a;=3,or  —S±:V^, 

37.  Given  a;' -f  a;'— 173;+ 16=0  to  find  x,    Ans,  a;=l,  3,  or  —6. 

38.  Given  4a;'-48a;»4-46a;— 11=0  to  find  x. 

Ans,  a;=i,  i,  or  11. 

39.  Given  2a;'  +  3a;»+a;— 30=0  to  find  a;. 

^Sdfe^r.v  ^ns,  a;=2,  or . 

40.  Given  6a;»  +  7a;'  +  39a;  +  63=0  to  find  x. 


Ans.  a;=  — li  orAdb^V- 251. 

41.  Given  a;"  +  6a;' -1600=0  to  find  x. 

Ans,  a;=10,  — 8±4*^"^. 

42.  Given  2a;»-33a;'  + 121a;- 84=0  to  find  x,  

Ans.  a;=4,  or . 

4 

43.  Given  a;»  +  6a;'— 3920=0  to  find  a?. 

Ans.  a;=14,  or  —lOdcQV^, 

44.  Given  2a;'— 9a;' + 9a;— 308=0  to  find  x. 

.  .  -5=M/^^^32Y 

Ans.  x=i,  or . 

4 

45*  Given  a;"- 29a;' + 198a;— 360=0  to  find  x. 

Ans.  a; =3,  6,  or  20. 
46.  Given  2a;'  +  9a;'  +  9a; — 308 = 0  to  find  x. 


.  ^         -11dtzV-^S2l 

Ans,  a; =4,  or . 

4 

47.  Given  4a;»— 112a;' +  109a;— 27=0  to  find  x. 

Ans.  a;=»,  I,  or  21, 

48.  Given  8a;'  +  34a;'  — t9a;4-30=0  to  find  x. 

Ans.  a;=i  li,  or  —6. 

49.  Given  4a;'-152a;'-f  149a;— 37=0  to  find  x. 

Ans.  a;=i  |-,  or  37. 

50.  Given  a;'— 15a;' +  74a;— 120=0  to  find  x,       Ans,  4,  6,  or  6. 


t 


CUBIC   EQUATIONS.  319 

'^  (337*)  The  solution  of  the  following  literal  equation  will  show 
that  there  always  is  a  binomial  square,  which  being  added  to  both 
members  of  a  cubic  equation,  after  it  has  been  multiplied  by  x,  that 
will  render  both  sides  perfect  squares.  The  only  difficulty  consists  in 
finding  the  binomial  square. 

It  will  be  perceived  by  looking  at  the  values  of  the  unknown  quan- 
tity in  the  preceding  examples,  that  this  mode  of  solution  is  practic- 
able when  one  or  more  of  the  values  of  the  unknown  quantity  is  a 
whole  number,  and  also,  when  one  or  more  of  the  values  is  a  mixed 
number,  or  a  proper  fraction. 

PROBLEM. 

Given  x*—(a-\-b  +  c)x^-\-  {ah  -\-ac-\-  bc)x — abc = 0  to  find  the  values 
of  a?. 

SOLUTION. 

4x*—^a  +  b-{-c)x^  +  4{ab-{-ac-{-bc)x'-^4tabc=0, 
[2x'  —  {a-{-b-{'c)xY—{a'—2ab—2ac-\-b''—2bc  +  c^)x^—4abc=:0, 
[2a:'— (a  +  &  +  c)xY  +  2&c[2a:'  -  (a  +6  +  c)a;]  +  6V=(a-6— c)V-^- 
2^a— 5— c)a;+6V, 
2x'-^{a+b  +  c)x  +  bc=(a—b—c)x  +  bc,  or  —(a^b—c)x—bCy 
.'.  2x=2a;oT2x^—2{b  +  c)x=:^2bc, 

b-hc±(b-c)     . 

x=a  x= i^ ^=0,  or  c. 

2  ' 

Remark. — An  inspection  of  this  example  will  enable  the  student  to  see  what 
relation  exists  between  the  roots  of  the  equation  and  the  added  binomial  square, 

QUESTIONS. 

1.  What  number  is  it  whose  third  part  multiplied  by  its  square 
gives  1944?  Ans,  18. 

2.  What  number  is  it  whose  ^,  i,  and  |,  multiplied  together,  and 
the  product  increased  by  32,  gives  4640  ?  Ans.  48. 

3.  There  is  a  number  such,  that  its  4th  power  divided  by  the  8th 
part  of  the  number,  and  167  subtracted  from  the  quotient,  the  re- 
mainder is  12000.     What  number  is  it  ?  Ans.  11^. 

4.  Some  merchants  engage  in  business  ;  each  gives  to  it  1000 
times  as  many  dollars  as  there  are  partners.  They  gain  in  the  busi- 
ness $2560  ;  and  it  is  found  that  they  have  gained  exactly  half  their 
own  number  per  cent.     How  many  merchants  are  there  ?     Ans.  8. 


320  CUBIC    EQUATIONS. 

6.  A  capitalist  puts  out  $10000  at  interest,  and  adds  the  interest 
yearly  to  the  capital.  At  the  end  of  the  third  year  he  finds  his  capi- 
tal increased  to  $11 5*7  6 1.     How  much  per  cent,  did  he  receive  ? 

Ans.  5  per  cent. 

6.  There  are  two  numbers  whose  difference  is  4,  and,  moreover,  are 
such  that  their  product  multipHed  by  their  sum  gives  1386.  What 
numbers  are  they  ?  Ans.  1  and  11. 

7.  Some  oflficers  were  in  a  field  with  a  detachment,  partly  infantry 
and  partly  cavaliy.  Each  oflScer  had  under  his  command  3  times  as 
many  cavalry  and  Y  times  as  many  foot  as  there  are  oflScers.  Each 
cavalry  soldier  has  2,  and  each  foot  soldier  22  cartridges  more  than 
there  are  oflScers.  They  had  altogether  15360  cartridges.  How 
many  oflScers  were  there  ?  Ans.    8. 

8.  A  person  being  asked  how  much  he  had  expended  that  day, 
answered,  ^'  I  have  spent  $4  more,  and  yesterday  twice  as  much  as  I 
did  the  day  before  yesterday  ;  if  I  multiply  together  the  sums  which 
I  expended  in  dollars  during  these  three  days,  and  add  756  to  the 
product,  I  obtain  exactly  134  times  as  much  as  I  have  expended  to- 
day."    How  much  did  he  expend  that  day  ?      Ans.  6  or  9  dollars. 

9.  Some  merchants  jointly  form  a  certain  capital,  in  such  a  way 
that  each  contributes  10  times  as  many  dollars  as  they  are  in  num'ber ; 
they  trade  with  this  capital,  and  gain  as  many  dollars  per  cent,  as 
exceed  the  number  of  merchants  by  8.  Their  profit  amounts  to  $288. 
How  many  were  there  of  them  ?  Ans.  12. 

10.  Some  merchants  collect  a  capital  of  $8240.  To  this  each  con- 
tributes 40  times  as  many  dollars  as  there  are  of  them.  With  this 
whole  sum  they  gain  as  many  dollars  per  cent,  as  there  are  persons. 
They  then  divide  the  profit  ;  and  each  takes  10  times  as  many  dollars 
as  there  are  persons  ;  but  after  this  there  remains  $224.  How  many 
merchants  were  there  ?  Ans.  Either  7,  8,  or  10. 

11.  The  3d  power  of  a  number  added  to  9  times  its  2d  power,  27 
times  the  number  and  27  more,  is  equal  to  125.  What  is  the  num- 
ber? Ans.  2. 

12.  If  the  difference  of  two  numbers  be  multiplied  by  the  2d 
power  of  the  greater,  and  the  sum  of  the  two  numbers  by  the  2d 
power  of  the  greater,  the  sum  of  the  two  products  will  be  432  ;  and 
the  difference  of  the  products  280.     What  are  the  numbers  ? 

Ans.  6  and  3f . 


CUBIC  EQUATIONS.  821 

18.-^4  and  B  commenced  to  speculate,  each  with  the  same  sum 
of  money ;  after  a  certain  number  of  months,  A  had  the  3d  power  of 
the  number  of  dollars  which  he  had  at  first,  wanting  36  times  its  2d 
power.  B  had  432  times  as  much  as  he  had  at  first,  wanting  $1V28 ; 
then  the  sum  of  what  A  and  B  had  was  equal  to  $343.  What  sum 
had  A  and  B  at  first  ?  Ans.  $19. 

14.  There  are  three  numbers ;  the  second  is  2,  and  the  third  3 
more  than  the  first,  and  their  continued  product  is  40.  What  are 
the  numbers  ?  Ans.  2,  4,  and  5. 

lb.  A  has  $4  more  than  B  ;  but,  if  the  number  A  has  be  multi- 
phed  by  the  2d  power  of  the  number  B  has,  the  product  will  be  225. 
What  number  of  dollars  has  each  ?  Ans.  A%^  and  B  $5. 

16.  *  The  sum  of  two  numbers  is  10,  and  the  difference  of  their  4th 
powers  is  1040.     What  are  the  numbers  ?  Ans.  6  and  4. 

17.  The  difference  of  2  numbers  is  8,  and  the  difference  of  their 
4th  powers  is  14560.     What  are  the  numbers?       Ans.  11  and  3. 

18.  The  joint  capital  of  3  partners  was  $66000.  ^'s  money  was 
in  trade  3  months,  ^'s  4  months,  and  (7's  5  months.  When  they 
shared  stock  and  gain  A  received  $3900,  B  received  $2700,  and  (7's 
gain  was  $750.     What  was  each  partner's  share  of  the  stock  ? 

Ans. 


*  Note — Let  a;  +  y  and  x—y  represent  the  numbers.  This  mode  of  repre- 
sentation frequently  produces  equations  simpler  than  those  produced  by  the 
usual  representation. 

21 


CHAPTER    XIII.. 
BiaUADBATIC    EaUATIONS. 

(3 3  8.)  Biquadratic    Equations   may  be   divided    into    three 
classes;  namely, 

Pure  Biquadratics; 
Incomplete  Biquadratics;  and 
Affected  Biquadratics. 

339*)  A  Pure  Biquadratic  is  an  equation  in  which  the  unknown 
quantity  is  contained  in  but  one  term,  and  its  exponent  is  4,  or  a 
fraction  whose  numerator  is  4  and  whose  denominator  is  an  odd 
number;  as, 

aa;* = ±5 ;  cx^=-±ih\  bx^  =  ±c,  <fec. 

PROBLEM. 

(340,)  Given  a;*=l  to  find  the  value  of  a?. 

SOLUTION     1. 

ar'=±l, 

.-.    a?=:l, -1,1/^,  or -V^. 
solution    2. 

«*— 1=0, 

(a:'-l)(a:'4-l)=0, 

whence,    «*— 1=0, 

and    ic*  4-1=0, 

.-.    a;'=l, 

and     a;'=--l, 

whence,    aj=l,  or  —1, 

and     x=V^,  or  -Y^. 


BIQUADRATIC  EQUATIONS.  323 

SOLUTIONS. 

a;*-l=0, 
(x-l){x'-}-x'  +  x  +  l)=0, 
whence,     a;— 1=0, 
and     x^  +  x^+x-\-l=:0  (a) 

X=:l, 

Equation  (a)  is  a  cubic,  and  is  solved  thus, 

4x*  +  4a;'  +  4a;'  +  4a;=0  (^)=(«)  x  4a?. 

(2a;''  +  a;)''+3.T'  +  4a:=0, 
(2aj'+a;)'  +  2(2a;''  +  a;)  +  l=a;''-2a;+l, 

2a;''  +  a;4-l=a;— 1,  or  1— a;, 
.  • .     2a;''=  — 2  ;  or  2a;'=  —2a?, 
♦  a;'=~l  2a;=— 2, 

x=±:V'^       x——!. 

SOLUTION      4.  ♦ 

a;'4-a;'+a;  +  l=0, 

a;»(a;  +  l)  +  (a;  +  l)=0, 

(a;4-l)(a;''  +  l)=0, 

whence,     a; +1=0, 

and     a;' +  1=0, 

.-.     ar=-l, 

and     a;'  =— 1, 

X  =  ±iV~^l. 
EXAMPLES. 

1,  Given  a?*=16  to  find  x.  Ans.  a;=db2,  ±2f^~l. 

2,  Given  3a;*=243  to  find  x.  Ans.  a;=±3,  ±:dV—l. 

3,  Given  a;*=256  to  find  x.  Ans.  a;=±4,  d=4f — 1. 

4,  Given  x*=:a''  to  find  x,  Ans.  a;=±a,  zhaV^l. 

5,  Given  aj*= 6  to  find  a;.  Ans.  x=±:V±:b. 


6.  Given  aa;*=c  to  find  a;.  Ans,  a;=db|/db- 


324  BIQUADRATIC  EQUATI0:NS. 


AFFECTED  AND  INCOMPLETE  BIQUADRATICS. 

(341*)  An  Affected  Biquadratic  is  an  equation  in  which  the  un- 
known quantity  occurs  in  but  four  terms,  the  least  exponent  of  the 
unknown  quantity  being  1,  or  a  proper  fraction  whose  numerator  is 
1,  and  the  exponents  of  the  unknown  quantity  in  the  other  terms 
being  respectively  two,  three,  and  four  times  as  large ;  as, 

x^ -\-a^  -\-hx^ ■\-cx^=d  \  oi? -\-ax\\-hx-{-cxk^d^  <fec. 

(342.)  An  affected  biquadratic  becomes  an  incomplete  biquadratic 
by  omitting  any  one  or  two  of  the  terms  which  contain  the  unknown 
quantity,  after  the  first  term,  considering  the  equation  to  be  arranged 
so  that  the  exponents  form  a  descending  series,  and  the  first  term  to 
be  that  which  has  the  greatest  exponent,  and  provided  that  when  the 
exponents  of  the  unknown  quantity  in  the  second  and  fourth  terms 
are  fractions  whose  denominators  are  even  numbers,  these  terms  shall 
not  be  omitted.     Thus, 

x*-\-ax^-\-bx'^=d  '^   X*' -\- ax^ -\- cx=id  \   x* -^bx^-{-cx=d  ;   x*  +  ax^=d; 
x*  +  cx=zd\  x*-^bx^=d. 

3.  3.  JL  1.  3. 

x'-^-ax^  +bx=d',  x^-i-ax^+cx^=d;  x''-{-bx  +  cx^=d;  a;'+aa;2=c?; 

x^-\-cx^=zd;  x^+bx'^=d, 
are  incomplete  quadratics. 

(343.)  An  incomplete  equation  of  the  form  ax*zt.bx'=±e,  or 

A  J.  .  .  . 

axn±:bxn  =  zt:Cjn  being  an  odd  number,  is  susceptible  of  a  general 

solution, 

PROBLEM. 

Given  ax*  ■\-b3^=c  to  find  the  values  of  x, 

SOLUTION. 

ax*-\-bx^=zc, 
Aa^x*  +  4abx'' = 4ac^ 


ir'  — 

dcVb^'  +  ^ac, 
-b±iVb'^4.ac, 

2a 

«=±- 

^/-6±f/6»  +  4ac 

2a 


BIQUADRATIC  EQUATIONS.  825 

{344»)  It  is  sometimes  advisable  to  consider  two  or  more  terms 
one. 

PROBLEM. 


Given  9x—4x'+  'V'4a;''— 9ar  +  ll=5  to  find  the  values  of  a?. 

SOLUTION. 


4x''—9x—V4x^—9x  +  U  =  '-5, 
4a;'— 9a;  +  ll  — i^4a;"— 9a;  +  ll=6. 


Considering  ^4x^ — 9aj  + 11  as  one  term,  and  putting  it  equal  to  y, 
the  equation  becomes 

y=——=8,OT-2, 

.*.    4«"— 9a;  +  ll=y'=9,  or4, 

4«*-9a:=  — 2,  or  -Y, 


9^7         9±V--31 


9db4/-31 
x=:2,  or  1 ;  or ^ , 


EXAMPLES. 

1*  Given  a;*  +  4a;''=ll7  to  find  x,  Ans,  x=S, 

2.  Given  x^ +  1x^=4:4:  to  find  x.         Ans.  a;=db8,  or  (—11)^. 

3,  Given  X*—I4x^  =  —  1225  to  find  x.        Ans.  a;=d=5,  or  ±7. 
4«  Given  «*— 6a;''=2'7  to  find  x.  Ans.  a;=±3,  or  ±  f^— 3. 


5.  Given  a:"'' 4-11  + Va?' 4- 11  =42  to  find  a;.      * 

^W5.  a?=±5,  or  ±1/38. 


6.  Given  a:*— 7a?  4-  Va;'— 7a?  4- 18=24  to  find  x. 

Ans.  x=:9,  —2,  or . 


7,  Given  x*  4-  f  6a;  +  a;" =42— 5a?  lo  find  x. 

.  ^        ^        -5±V22i 

Ans.  a; =4,  —9,  or , 

'  2 

8.  Given  (a?'4-5)'-4ar'  =  160  to  find  x. 


Ans.  cc=dt3,  or  dri^— 15. 


826  BIQUADRATIC  EQUATIONS. 

1  2 

10.  Given  a;*— 2ic  +  6V^a;''— 2a;  +  5  =  ll  to  find  ar.    ,  __ 

Ans.  x=l,  1  or  1±2V15. 

11,  Given  2ar'  +  3a?— 5|/2a:'  +  3ar  +  9  =  — 3  to  find  x.  

o        .,        -3±|/-66 
Ans.  x=zS,  —41,  or . 


12,  Given  9x-{-  f^  1 6a;'' +  3 6a;' = 15a;* -4   to  find  a;.  

9±f/481 
Ans.  x=l^,  -i,  or 


50 


(8\  8 

a;H — I  +a;=42 —   to  find  a;, 
a;/  a; 


^715.  a;=2,  4,  or  , 

14.  Given  a;(f^5+l)'=102(a;+V''i)-2576  to  find  a;.  _^ 

93  =F  4/186 


^w«.  a;=49,  64,  or 


2 


15.  Given  6ar'+2V9a;''— 24a;=16a;  +  12  to  find  a;.  _ 

^        ^        4±2fl3 
Ans.  a;=3,  -i  or . 


,6.  Given  J^±^J±±^£+^+^l)  ^  fi^^x. 

3  4/a;'+a;  +  6 

-lrbV/377 
^Tis.  a;=5,  —6,  or , 

17.  Given  (ar*— l)(a;'  — 2)4-(a;'— 3)(a;''  — 4)=a;*  +  6  to  find  x. 

Ans.  a;=±l,  or  ±3. 

18.  Given —  a 1 =0  to  find  x. 

a;«+lla;-8^  a;''  +  2a;-8^a;=-13a;-8 

Ans.  a;==bl,  or  ±8. 
The  resulting  equation  in  this  problem  after  reduction  is 

a;*— 65a;'=  — 64. 
In    solving    this,  to    avoid    large    numbers    put    65= 2a,    then 
— 64=  — 2a  + 1.     The  equation  after  substituting  becomes  : 
a;*— 2aa;''=— 2a  +  l, 
«* — 200^  H- a' =a' — 2a  + 1 , 

a;"— a=a— 1,  or  1,  —a. 
3^^=20— 1;     or  0^=1, 
a;«=64  x=±l, 

aj==t:8. 


BIQUADRATIC  EQUATIONS.  827 

(345.)  Biquadratic  equations  may  be  reduced  to  simpler  forms 
when  both  sides  are  perfect  powers,  zero  being  considered  a  perfect 
power,  by  extracting  the  root.  Sometimes  artifice  is  necessary  to  get 
the  equation  in  a  proper  form  for  reducing.  The  artifices  that  may 
be  employed  are  numerous ;  particular  ones  being  applicable  only  to 
particular  problems. 

P  R  O  B  IlE  M 

1.  Given  a;*— 4a;'  +  6a;'— 4aj+l=0  to  find  the  values  of  a?, 

S  OLUTION    1 . 

Taking  square  root  twice,  or  the  4th  root,  we  have 

a;— l  =  ±y±0=0,  — 0,  |/^,  or  -^"^=0,  0,  0,op  0. 
x=l,  1, 1,  or  1. 

SOLUTION     2  . 

a;*— 4a;' +  6a;'— 4a;  + 1=0, 

(a;«-i)(a;_l)(a;-l)(a;-l)=0, 

whence,  a;— 1=0, 

a;- 1=0, 

a;-l=0, 

a;- 1=0, 

a;=l,  1,  1,  OP  1. 

PROBLEM 

2.  Given  a;*— 4a;'+6a;'— 4a;'4-l=6  to  find  the  values  of  a. 

SOLUTION. 

a;*— 4a;»  +  6a;'— 4a;  +  l=6. 
a?-l  =  ±4/±6^ 

a;=l±|/±6,  

a;=l±i/6,  or  lil/— 6. 
Or, 
«"-2a;  +  l  =  ±V6, 
a;— 1  =  ±V±6, 
a;=ld=l/db6. 

PROBLEM 

49«P»     48  6 

8.  Given  -\ — r— 49=9-f-  to  find  the  values  of  a?. 

4        ar  X 


BIQUADRATIC  EQUATIONS. 
SOLUTION. 

49a;'     48     ,„     ^     6 

+  49=9+- 

4         a;  X 

49a;'     .^  .  48     ^     6 
—--49+— =9  +  -, 

4905''      ,^     49     ^  .6     1 

49  +— =9+-  +  -, 

4  a;  a;    a; 

"^aj    '^     o     1  „     1 

— =3  +  -  or  -3 — , 

2     a;  a;  x 

Ta;    „     8  7aj        „     6 

—-=3  +  -;  or  —-=  —  3+-, 

2  x'  2  a;' 

7a;'=6a;+16  'ra;''=-6a;+12, 

Yaj'— 6a;=16  7a;'+6a;=12,  _ 

3±11     „           ^,                         -3±i^93 
a:=— y— =2,  or  -1|  a;= , 


PROBLEM 


4.  Given  /(aJ-2)''-a;)y-(a;-2)»=88-(a;-2)  to  finda;. 


SOLUTION 


/(a;-2)''-a:)V-(a;-2y=88-(a;-2,) 

/(a._2)'~arV-/(ar-2)'-ar)\=90, 

1±19 
(a;~2r-a;=-2— =10,or-9, 

a:'— 5a;=6  ;  or,  a;'— 6a;  =  —  18, 

6±7     ^  ,  5±3V-8 

irrr-— — =6,  or  —  1  ar= 

2  2 

PRQBLBM 

^    _.  12  +  8ar^       ^    , 

6.  Given  aj= to  find  x, 

a;— 5 

SOLUTION. 

12  +  8ari 


ar= 


x—^    ' 

«»--6a;=12  +  8ar^, 
a;*— 4a;=12  +  8ari+aj, 
a?*— 4ar  +  4  =  1 6  +  8a?i  +  ar. 

ar— 2=4+ari,  or  — 4— ari,  pbrwardi 


BIQUADRATIC  EQUATIONS.  829 

x—xi=6;  or  a?  +  ari=  — 2       

ari= =3,  or  —2  a?i= r 

2  2 

a;=9,  or  4  «= r 


EXAMPLES. 

1,  Given  a;*— 8a;'+24a;*— 32a;+ 16=81  to  find  x.  

Ans.  x=5,  —1,  or  2±3i^— 1. 

2i  Given  a;*— 8a;''+24a;''— 32a;=240  to  find  x. 

Ans.  a;=6,  —2,  or  2±4V^. 

3*  Given  —  H — l7a;=:8  to  find  x. 

2        4 


Ans,  fl;==fc2,  or  —J,  or  —8. 


4.  Given  27^-^+^=^-i-,-f  5  to  find  x, 
3x^      3       3a;      3a:' 


^ws.  ar==2,  — 1|,  or -. 


5»  Given  a;' — :r+i5=-^r:; rs  to  find  x, 


5x     .        25a:''     64 

2-^^^=l[6~^  

-2±21/-71 

Ans,  x=4,  —8,  or • 

y 

*    ^.  a:*  6  351   ^    ^    , 

6*  Given  7-5 — 7r4--i — T=-iTr-5  ^  ^^^  ^• 

(a;''— 4)      a:"— 4      25a:'  

Ans,  x=z±:S,  or  zhV'ff. 

7t  Given  a; +  4—24/—^= to  find  x, 

^  x—4t     a;--4 

^n«.  a;=db6,  or  dbVlY. 
8.  Given  3((a;--l)'~a:W2a:=34l4-2(a;-l)'  to  find  x, 

3|/3±1^-109 
^w».  a;=5,  —2,  or 


21^3 


X  4         21 

9*  Given -H —         = —  to  find  x. 

x  +  4:    ^a;  +  4      ^ 


-.«        o        49±f/3185 
^rw.  a;=12,  —3,  or • 


10.  Givena;'— 2=2V'6— 4a;tofinda;.  _  

Ans.  x=—ldbV5,  or  Izt^^S. 


830  BIQUADRATIC  EQUATIONS. 


x4-  vx' Q 

11,  Given     ^    =(x-2y  to  find  x, 

x-Vx'-9 


Ans,  x=Sj  6,  or  . 


12.  Given  ^^     +—777 — =—  to  find  a?. 

i/5x'-x'        25a;         ar  

^TOS.  ar=±2,  or  ±1/— '720796. 

13.  Given  (a;4-6)'+2a;i(a;  +  6)  =  138  +  a:i  to  find  x. 

Ans.  x=4:,  9,  or -2- . 

2 

II.  Given  (a:— 2)''— 6ari(a;— 2)  =  24~14a;  +  16a^  to  find  a;. 


'^  Ans.  x=l,  16,  or . 

'      '  2 

15.  Given  (4a;+l)'  +  4a:i(4a;+l)  =  1912  — (10a;  +  3a;i)  to  find  x. 

Ans.  .=9,  121,  or  =^2±^El^, 
8 

273;  — 

16.  Given  a;' _  4. 25 = 7  4/a:(5  —a;)  to  find  x.  4 

209qil3f/249 


^715.  a;=4,  65^,  or 


8 


3a; 


17.  Given  8a?»— 13=— +i/6a;''+52a;'  to  find  x. 


A  o       ,.        3±|/3337 

^w«.  aj=2,  —If,  or 


64 


4a:» 


18.  Given  4a;'  +  21a?  +  8ar*t/7a;'— 6a-=207 to  find  x. 

3 


A              o   .ni«         -129±3V-2567 
Ans.  a;=3,  12if,  or . 

19.  Given  ar*— 8a;»  — 12a;»  +  84a;=63  to  find  x. 

Ans.  x=:2doV1±:  j/udhV^. 

8-?- 

20.  Given  -— -—= to  find  x. 

6  a:''  3 

Ans.  x=z4,  —2,  or  —  Idr^^^. 

21.  Given  a:*  — 1 2a: =5  to  find  a;,    ^tw.  3^=1=1:^/2,  or —l=b  2  V^. 

4Yx 48  — 

22.  Given  x= — -  to  find  x.        Ans.  x=4,  16,  or  8q:2|^7. 

3/ ~"~  1  o 


BIQUADRATIC  EQUATIONS.  381 

23.  Given  ar-4— ^=5(1+^)  to  find  x. 

^x      X        \       xj 

Ans,  a; =4,  9,  or . 

24.  Givena;*— 25a;''  +  60a;— 36=0tofinda;. 

Ans.  x=\,  2,  3,  or  —6. 

25.  Given  a;*— 36a;' +  720;— 36=0  to  find  a;.        _  __ 

Ans.  a:=:3±l/3,or  -3±n6. 

26.  Givena;*— 6a;'  — 8a;-3=0tofinda:. 

Ans.  x=Z,  —1,  —1,  or  —1.^ 

27.  Given  a;*— 9a;''+4a;4-12=0  to  find  x. 

Ans.  a;=2,  2,  —1,  or  —3.     • 

28.  Givena;*— 6a;'  +  24a;— 16=0tofinda;. 

Ans.  a;=±2,  or3±4/6. 

29.  Given  a;*— '7a;»+59Ja;-'72i=:0  to  find  x.  __ 

An8,.x-H-±Wl5> 

30«  Givena;*— iVa;'— 20a;-6=0tofinda;.        _  

^ .,  Ans.  a;=2db|/7,  or  -2^1^-2. 

31.  Givena;*— 65a;'— 30a;+604=0tofinda;. 

Ans.  a:=3,  Y,  —4,  or  —6. 

32.  Givena;*— 3a;'— 4a;— 3=0tofinda;.         _  

1±V13         -l±i/-3 
Ans.  X—  — ,  or . 

,   33.  Givena;*— 27a;'  +  14a;  +  120=:0  tofinda:. 

Ans.  a;=3,  4,  —2,  or  —6. 

34.  Given  a;*+6a;'— 24a;— 16=0  to  find  x. 

Ans.  a;=2,  —2,  or  — 3±|/5. 

35.  Given  a;*  -  45a;'  -  40a;  +  84 = 0  to  find  x. 

Ans.  a;=l,  1,  —2,  or  —6. 

6    rT       3    /^      3 v/a;' 

36.  Given  4/  -,  +  4/  -= to  find  x. 

f   X*      ^   X  X  21 

Ans.  a;=itl,  or  ^-5-- 


8 


37.  Given  a;*/l  +^)  _(3a;'  +  a;)  =  70  to  find  x. 


— IdbV— 261 
Ans.  a;=3,  —3^,  or , 


^2  BIQUADRATIC  EQUATIONS. 


88«  Given  ^       H = =— -  to  find  x. 

Vx*  -9x'         ^^         2a; 


Ans.  a;=±6,  or  ±^^'16661. 


39.  Given  x^—2x+ 4=2^x^  —  1  to  find  «?.  _  

Ans.  x=4:±:  ^/6,  or  ±  1^—2. 


40.  Given  aj— 2V'a;  +  2=l+Vic'— 3£c4-2  to  find  a;.  ^ 

Ans,  aj=9±4r7,  or  — - —  . 

(346.)  An  equation  of  the  form  ax*zthx^dzcx'±:bx  +  a=0,  is 
called  a  recurring  equation  of  the  fourth  degree,  or  a  biquadratic 
recurring  equation, 

PBOBLEM. 

(347.)  Given  ax*  +  bx''+c3^  +  bx-{-a=z0  (1)  to  find  the  values 
of  a?. 

SOL  UTION. 
By  multiplying  by  4a  we  get 

4a^x*  +  4abx^  +  4acx''  +  4ahx+W=zO  (^)=(1)  X  4a. 

{2ax'+bxY  +  (4ac-  b')x^  4-  4abx  +  4a''=0  (-B) 

(2aa;''+6a;)'*  +  4a(2aa;''  +  6a;)  +4a''=(8a''  +  6'-4ac)a;»  (0) 
2ax^  +  bx+2a=dixV8a^  +  b^—'iac 


2aa;'  +  (6:f  j/8a"+6'— 4ac)rc=:— 2a 


ifSa'+S— 4ac— 6±i/— 8a'  +  26''— 4ac±26V'8a'+6'— 4atf 

a?= :; . 

4a 

It  may  be  perceived  that  the  coefficient  of  £c'  in  the  term  that  is 
added  to  both  members  of  (A),  and  which  makes  the  first  member  a 
perfect  square,  is  a  function  of  known  terms. 

This  coefficient  is  equal  to  the  coefficient  of  «*  in  (B)  subtracted 
from  twice  the  coefficient  x*  in  (A),  or  confining  the  explanation  lo 
the  primitive  equation,  is  equal  to  8  times  the  square  of  the  coefficient 
of  x*j  plus  the  square  of  the  coefficient  of  a;',  minus  4  times  the  pro- 
duct of  the  coefficient  of  x*  by  the  coefficient  of  a;^ 

If  the  primitive  equation  had  been  multiplied  by  a  instead  of  4a, 
the  coefficient  of  x^  in  the  term  added  would  have  been  just  J  as  much, 


or 


BIQUADRATIC  EQUATIONS.  333 

Node. — This  mode  of  treating  recurring  equations  of  the  fourth  degree  is 
original.  The  method,  it  will  be  perceived,  is  a  general  one.  The  discovery 
was  afterward  independently  made  by  M.  C.  Stevens,  who>  at  our  request,  has 
furnished  the  following  concisely  written  rule.  In  applying  it,  the  original 
equation*  must  not  be  multiplied  by  4,  as  was  done  in  our  solution,  in  order  to 
avoid  fractions. 

RULE. 

Divide  by  the  coefficient  of  c»*  and  transpose  the  term  containing  a?» ;  then  a^dd  to 
each  memlefr  2x'^+  the  square  of  half  the  term  containing  x,  or  —27?+  the  same, 
according  as  the  second  and  the  fourth  terms  have  like  or  unlike  signs.  Extract  the 
square  root  and  the  equation  reduces  to  the  second  degree. 

(348.)  The  following  is  the  plan  usually  given  for  the  solution  of 
a  recurring  equation  of  the  fourth  degree  : 

PROBLEM. 
Given  ax*-\-hx^-{-cx^-\-hX'\-a=iO  to  find  x, 

SOLUTION, 

oaj'  +  Ja^  +  c-l— +  ^=0 
X    a?" 

«     a      ,       h 

aar*+-5  +  6a;-|--+c=0 
x^  X 

Now  put  x-\ — =iy 

X 

Then  x^+-=y^—2 

X 

Substituting  a(y'*  —  2)+6y4-c=0 


yz=. ! 

X'\ —  =  — _ 

X  2« 


2aar' +  (Jq^l^Sa'' +  5'— 4ac)a:=  — 2a 

±|/8a'  +  6'— 4ac  -  6  ±  |/— 8a"  4-  26»  -  4ac  ±  264/8a''  +  6»— 4ac 


4a 


384  BIQUADRATIC   EQUATIONS. 

'  PROBLEM. 

Given  10a;*— 9a;'— 20a:''— 9a; +  10  =  0  to  find  the  values  of  «. 

SOLUTION     1. 

10a;*— 9a;'— 203;"— 9a;  4- 10=0, 

400a;* -SeOa;"  —  800a;'— 3  60a; +  400=0, 

(20a;''-9a;)''— 881a;'— 360a;  +  400=0, 

(20a;'— 9a;)'+40(20a;'— 9a;)+400=1681a;'. 

20a;'— 9a;+20=±41a;, 

.-.     20a;'-50ar=  — 20;       or  20a;'  +  32a;=— 20, 

2a;'— 5a;=— 2  5a;'  +  8a;=— 6, 

5±3     ^        ,                      -4±3f^i:T 
a;=:'-^=2,  or  ^  x= . 

SOLUTION      2. 

10a;*— 9a;'— 20a;'  — 9a;+ 10=0, 

10a;'-9a;'-20--  +  i^=0, 
X    aj' 


10(.'+J)-9(x+l)=20, 


put    «+-=y, 

then     a;«+-=:y'— 2, 
a;'     ^        ' 

substituting  lOy'— 20— 9y=20, 

10y'-9y=40, 

9±41      6  8 

^=-20r=T'^^-'5' 

.15  8 

.• .    x-i — =-,  or  — , 

a;     2'  5' 

and     2a;'— 5a;=— 2;  or  6a;' +  8a;  =—6, 

-5^3_                                     -4±3i^"=T 
— _2,or^  x= . 


x= 


EXAMPLES. 

1,  Given  a;*  +  24a;'— 114a;'— 24a; +  1=0  to  find  x. 

Ans.  x=2:±zV5,  or  —14±  1/197. 

2.  Given  a;*  +  6a;'  +  2a;' + 6a;  + 1  =0  to  find  x. 

Ans,  a;=±|/-l,  or  ~^'^^^\ 


BIQUADRATIC  EQUATIONS.                               385 
3*  Given  x*+x'-{-x^+x-\-l  =0  to  find  x.  . 

-     -h         -  -f  ^ 

4»  Given  2a;*— 4a;'— 6ar'— 4a;  +  2=0  to  find  ar. 


Ans,  x=: . 


5.  Given  3a;*  +  2a;'  +  43;'  +  2a; + 3 = 0  to  find  x. 


Ans,  x= . 

6.  Given  4a;*  +  3a;' — 8a;' — 3a;  +  4 = 0  to  find  x. 

■Z±V1% 


Ans.  a;=r±l,  or 


7.  Given  7-- — r.=*  to  find  x. 
(1  +  a;)* 


Ans.  a;=l=fcf/3dby3|/i/3db2. 

(349*)  There  are  other  biquadratics  that  are  not  recurring  which 
are  susceptible  of  a  similar  solution,  but  the  coefficient  of  a;'  must  be 
decided  by  trial. 

EXAMPLE^. 

1,  Givena;*— 2a;'— Ya;*— 8a;+16=0lofinda;. 

.             ^    ^        -3dti/3Y 
Ans.  a;=l,  4,  or . 

2,  Given  a;*-f »'— a;'-t-2a;4-4=0  to  find  x, 

A  -1^ 

Ans.  x= 

3,  Given  4a;*4-8a;»-89a;'4-28a;+49=0  to  find  x. 


-ldt:f/21±^-10qz2V'21 

Ans.  x= -2- , 

4 


Ans.  x=l,  3^,  or — . 

4.  Given  2a;*  +  24a;'-315a;'  +  216a;  +  162=0  to  find  a;. 


Ans.  x=—3dz^V94±:W^^=p^V94, 

5,  Given  a;*— 12a;'  +  4'7a;'— 72a;+36=0  to  find  x. 

Ans.  a;=l,  2,  3,  or  6. 

6.  Given  a;*— 9a;' +  15a;'— 2 'ra;  + 9=0  to  find  x.  

,              9doSV5±^1S±5W5 
Ans,  x= . 


886  BIQtJADEATiC   EQUATIONS. 

7.  Given  x*i-B6x^—400x^—Sl68x  +  1l44=0  to  find  x. 


Ans.  a;=  — 9±V/137±i/306q=18f/137. 

8.  Given  f^«--=  -:= to  find  x.   Ans.  a;=  1, 16,  or  '^^^-'^^ 

^    f^aj-2  2 

(350.)  There  is  a  class  of  problems  which  may  be  solved  after 
the  manner  given  in  the  solution  to  the  following 

PRO  B  LE  M. 

Given  a;*  +  2a;»— Yar*— 8a;=--12  to  find  the  value  of  a?. 

SOLUTION. 

x*-}-2x'  —  1x''—Sx=—12, 
(x^-\-xY—8x'—8x=^12, 
{x'  +  xy-8(x'  +  x)=^12, 
{x'-\-xy-8{x^  +  x)  +  16=4, 

x^  +  x—4=±2, 
x^-{-x=6^0T  2j 

a;=2,  —3,  1,  or  —2. 

EXAMPLES. 

!•  Given  x*-\-2x^—3x^—4x-{-4=0  to  find  x. 

Ans.  a?=l,  1,  —2,  or  —2. 

2.  Given  x*—l  2x^  +  60a;'  —84a;  +  49 = 0  to  find  x. 

Ans.  a;=3±j/2,  or  8±V'2. 

3.  Given  a;*— 10a;*  + 3 6a;'— 60a; +24=0  to  find  x. 

Ans.  x=l,  2,  3,  or  4. 

4.  Givena;*  +  2a;'— 13a;»— 14a;4-24=0  tofinda;. 

Ans.  ar=l,  3,  —2,  or  —4. 

5.  Given  x*  +  12a;'  +  64a;'  +  108a;  +  81  =0  to  find  x. 

Ans.  x=—S,  —3,  —3,  or  —3. 

6.  Given  x*  -{-2qx''  +  3q^x^  +  2q'x=zr*  to  find  x. 


Ans.  .--9±V-Sg^±^Vr*+q' 
2 

7.  Given  a;*— 14a;' +  6 la;'— 84a; +  36=0  to  find  x. 

Ans.  a;=l,  1,  6,  or  6. 

8.  Given  4a;*+|=4a;»  +  33  to  find  x. 


A             o       ^,        l±V^-43 
Ans.  x=2,  —1^,  or . 


BIQUADRATIC   EQUATIONS.  887 

9,  Given  a:*— 2a;'— 2a;'  +  3a;=108  to  find  x. 


Ans,  x=4:,  —3,  or . 


10.  Givena;*— 2a;'+a;=30  tofindar. 


Ans,  x=S,  —2,  or . 


11.  Given  a;*— 6a?»  +  6a;'  +  12a;=:60  to  find  x. 


Ans.  a;=6,  ~2,  or . 

'  2 

12.  Given  a:*— 8a:*  +  10a;' +  24a; +  6=0  to  find  x, 

Ans.  x=5,  —1,  or  ^zt^B, 

13.  Given  a;*— 2a;'4-a;=132  to  find  x. 

Ans.  a; =4,  —3,  or  ■ . 

2 

II.  Given  a;*— 2aa;»+(a*— 2)a;'  +  2aa;=a'  to  find  x. 


Ans.  xz=~db^^  +  l±Vl 


2^4 

15.  Given  a?*— Boa;' +  8a V  4-3  2a'a;= 9a*  to  find  x. 

Ans.  a;=2azhat/3,  or  2aitaV'13. 

-»    ^.        18     81-a;»       a;'-65      ^    , 

16.  Given  —  +  — = — to  find  x. 

x^         9x  12 

Ans.  a;=9,  —9,  —4,  or  —4. 

17.  Given  a;*— 2a;'— 25a;' + 26a; +120=0  to  find  x. 

Ans.  a;=3,  5,  —2,  or  —4. 

18.  Given  a;*— 12a;'  +  44a;'— 48a;=9009  to  find  x. 

Ans.  x=lS,  —7,  or  3±3f^— 10. 

19.  Given  ar'--2a;2  +^a;— |/.^z=6  to  find  x. 

Ans.  x=l,  4,  or . 

2 

20.  Givena;*— 6a;'  +  13a;»— i2a;=5  tofinda;. 

.              3±4/l3       3±i/^iri 
Ans.  x= — - — ,  or . 

21.  Given  a;*— 8aa;'  +  8aV  +  32a»a;=fl?  to  find  x. 


Ans.  a;=2a±4/8a'±t/l6a*+<?. 
22.  Given  a;*— 4a;' +  8a;'  — 8a; =21  to  find  x. 

Ans.  x=S,  —1,  or  liV'lIe. 
22 


B38  BIQUADRATIC  EQUATIONS. 

23.  Given  4a;*  +  24a;'  +  62a;'  +  48a;=:480  to  find  x. 


.  -3±|/l=t81/31 

Ans,  x= , 

2 

24,  Given  a;*+12a;''  +  40a;»  +  24a;=:837  to  find  x,  

Ans.  x=S,  —9,  or  — 3±i/— 22. 

25.  Given  ar*  +  6a;''  +  80a;'  +  213a;=2128  to  find_^^ 

— 3dbl/— 133±2Kr3553 

AtIS,   X=: . 

2 

(351*)  The  following  examples  are  best  solved  by  factoring,  since 
the  factors  are  readily  obtained.  The  solution  of  the  following  prob- 
lem will  serve  as  an  illustration. 

PROBLEM. 

Given  a;* -f  3a;' ~  3a; =9  to  find  the  values  of  a;. 

SOLUTIO  N . 

a*  +  3a;'— 3a; =9, 
^  a;*  +  3a;' = 3a; +9, 

(a;  +  3)a;'=3(a;  +  3), 
a;'=3, 

x=VS, 
Dividing  a;»-a  by  a:-'v/3,  we  get  it'-^-VSx +1/9=0, 

x^  +  VSx=—V9, 


_V3±f-3»N/9 

x=. — 


2 
and     a;4-3=0, 

a;=— 3. 


SXAllPLES, 

i3a;'  * 

1,  Given  x*-^ — 39a;=81  to  find  ar. 

3 


_L.«        -13=fct/-155 

Ans.  a;=±3,  or . 

6 

2.  Given  26a;* —624a;* =25  to  find  a;.  

Ans.  x=do5,  or  ±11/— 1. 

3.  Given  a;*— a;'— 2a;=4  to  find  x.  ^  

Ans.  x=2,  —1,  or  dbV—2. 


BIQUADRATIC  EQUATIONS.  339 

BIQUADRATIC  EQUATIONS  CONTAINING 
TWO  UNKNOWN  QUANTITIES. 
(352*)  In  eliminating  one  of  the  unknown  quantities,  the  student 
must  be  guided  by  the  methods  used  in  equations  of  the  lower  de- 
grees, adopting  that  mode  which  seems  to  be  best  suited  to  the  par- 
ticular problem  under  consideration.  In  the  following  examples  there 
are  some  which  belong  to  what  are  called 

HOMOaSNEOUS    EQUATIONS. 

(353.)  Homogeneous  Equations  of  the  fourth  degree  are  those 
in  which  each  of  the  literal  terms  contains  two  literal  factors ;  as 

for  these  equations  are  the  same  as  when  they  are  written  a;a?^'a;y=a 
and  yy-\-xx=h^  in  which  it  is  seen  that  each  literal  term  contains 
two  hteral  factors.  Homogeneous  equations  of  the  fourth  degree  are 
susceptible  of  a  complete  solution,  according  to  the  plan  exhibited  in 
the  solution  of  the  following 

PBOBLE  M. 


Gi^«=  {  ^+2^=65  \^^^='  ^""^  y- 


SOLUTION. 

x^-\-xy=5Q, 

a;y  +  2y''=60, 

Put  y=:nx    «' 4-^03*' =5 6, 


X': 


and    7mb'+-2%V=60, 
60 


n-h2n" 
56  60 

l  +  w~w  +  2»'" 
14  16 


l+n~n  +  27i'' 
28w»+147i=15-f  15w, 
28»'— w=15. 


I  or  -4. 


Whence,     a;==fc4|/2,  or  ±14, 
and  y=:db3i/2,  or  ±10. 


840 


BIQUADRATIC  EQUATIONS. 


Remark. — Some  of  the  following  examples  are  composed  of  homogeneous 
equations,  and  may,  therefore,  be  solved  after  the  manner  of  the  one  just  given. 
The  others  are  miscellaneous  in  their  character,  and  various  are  the  artifices  to 
be  employed  in  reducing  them.  The  student,  as  a  general  rule,  should  so  manage 
the  equation  as  to  obtain  a  perfect  square  in  the  left  hand  member,  and  this 
may  be  obtained  frequently  before  elimination. 

EXAMPLES. 

1.  Given  \  (^+^)' ''  (^-^)^  '-'-^^''^^l  to  find  x  and  y. 

Ans.  a;=:±9,  or  ±7,  y=±:*J^  or  ±9. 

35=  ±2,  or  iyiy, 
or  18. 


Ans. 


(y=3, 


^'^-A7:.7:f=iik\^'^'''-''- 


Ans, 


03= ±9,  or  ±4, 
±4,  or  ±9. 


{y= 


4.  Given  j^^^+^'f+r  '^  to  find  ^  and  y. 


Ans. 


9rt:i/-61 
a;=l,  3,  or , 


y=l,  3,  or 


9±|/-61 


5i  Given  * 


/^x-{-y 


11 


(^4-y)' 


y  4:Vx-\-y, 

x=f  +  2, 


to  find  x  and  y. 


a;=3,  6,  or 


Ans. 


y=l,  2,  or 


9±:3V--119, 

32 

—  3±t^— 119 
8  ' 


ly=±5, 


a;=±4,  <feo^ 
Arc. 


7.Giyen]"'+'+^=f-^[tofind.andy. 
(  ary=6,  ) 

ja;=:3,  2,  or  — 3rfc4^3, 

(y=2,  3,or-3=Fi/3. 


Ans. 


BIQUADRATIC   EQUATIONS. 


341 


8.  Given 


9«  Giyen 


1      -r     y  -ry  -r  y  '  t  to  find  a;  and  v. 

(                   xy-y'=z8,  f 

.  (  a:=6,  9,  or  — 9rf:f^5, 

"^'''*  1^=4,1,01-3^:1/5. 

1  .      .         on   f-  to  fi^d  rc  and  y. 

(       ajy+a;+y=39, )  ^ 


-4?15. 


-13dbV-39 
a;=9,  3,  or , 


-13zf|/-39 


10*  Given 


11,  Given 


12.  Given 


13,  Given 


14t  Given 


15 •  Given 


16.  Given 


y=3,  9,  or 

\    *^=f-^3'Htofind^andy. 

Ans.    P-'t>2,or3±m. 
(  y=2,  4,  or  3qpV21. 

\  ic=5, 1  or , 

Ans.    j  I7rp64/::^ 

Cy=3,  —15,  or  — 6±i/— 2. 
]         a;y=:2, )  ^ 

(y=l,  2,  -1,  or  -2. 
^  _  a  f   t<^  fijid  ^  and  y- 


icy: 


•s  ^  ic  +  y     '^     3aj         '>  to  find  a:  and  y. 
|aJ2^-(a;+y)=54,  ) 

^,...    ■i^=  6, -4X6,  or -4J, 
(y=12,  -9,  12,  or -9. 
(6a:'4-2y'=5ary+12,  )    ^    ^    ,  . 

1  «      .  o  a     o  3     ^  M-  to  find  re  and  y. 

Ans    i^=±2,&c., 
^^^-    |y=i3,&c 

C  a;V+«5y'=30,  ^ 

<      1     1     _       V   to  find  .r  and  y. 

i  ^^v  -"  S 

^"*-U=2,  3.  -e.orl. 


842 


BIQUADRATIC  EQUATIONS. 


17.  Given  }  f'^x^y-''^      x^' 


to  find  X  and  y. 


j  a:=4,  -2,  or  lil^A^ 


18.  Oive.  j  (;;;,)|;r^^-;^       [  to  and  .  and  y, 

j  a:=ll,  -1,  or  eii^^-STld, 
^**    (  y=l,  -11,  or  _61±4/-3Y16. 

19.  Given  J-StVUo^-Oy-ie.^.,  ?  ^fi,^,,,,^. 

(  6a?=4  +  25y^    ) 

j  a;=l,  1,  -/o,  or  -^^,      

20.  Given  j  "•"^■'^^Zjg^  f  to  find  .  and  ,. 


Ans. 


a;=5,  4,  or  8dbif/-Ai«, 
or  8TJ4/^^^^. 


j  ^=5,  4, 
(^=4,6, 


y+ 


21.  Given 


"a:      X 


x^       X  __64 
^3     2»/y     y\ 


to  find  a;  and  y. 


An,    i  ^=*' -WS  N>  O' W. 


^        ^   ,          -97±  1/6045 
aj=6,  J,  or ^^^ , 


Ans,   ^ 


y=6, 160,  or 


58 
1682 


9'7=pf6045 


^'<^^^Ay'-Z=Zr'^\-^''-'y 


(  a;=400,  225,  or  12i(— l-5f^— 23, 

^^**    \        .r.r.       OH.         125 ±25*/^=^ 
i  y=500,  — 3*75,  or . 


BIQUADRATIC   EQUATIONS. 


343 


24*  Given  \    .      *     ,  L  f  to  find  x  and  y. 


Ans. 


^=2,  1, ^— , 


25.  Given  ]  'ft'^'rf '.-^f;'  ^,,    [  to  find.  and,. 


-4»5. 


171        65=F^1114 
a?=3,.— -— -,  or 


133' 


26 


34  — 9±3'^/lll4 


^'^''m'"^ 


26 


(354*)  Biquadratic  equations  which  do  not  admit  of  solution  by 
any  of  the  previous  methods  may  frequently  be  solved  as  cubics  by 
the  addition  of  a  binomial  squared  to  both  members. 

PROBLEM. 

Given  a;*  +  4a;'— a;''--16a;=12  to  find  the  values  of  x, 

SOLUTION. 

a;*  +  4a;'-a;''-16a?=:12 

(a;''  +  2a;)''-6a;'-16a;=12 

(aj»4-2a:)'— 4(a;'  +  2a!)+4=a?'  +  8a;4-16. 

a?"  +  2a;— 2=a:  +  4,  or  — a;--4, 

.*.     a;^4-a;=:6  ;  or  3;"  + 3a;  =—2 

--1='=5     ^      „  -3±1         ,       ^ 

x= — - — =2,-3      x=: — - — =— 1,  — 2 


EXAMPLES. 


1,  Given  a;*— 6aj'  +  12ar'— 10a;  +  3=0  to  find  x. 

Ans,  x—ly  1,  1,  or  3. 

2,  Given  a;*-~4a;'  —  19a;'  +  46a;  + 120=0  to  find  x. 


Ans.  a;=4,  5,  —2,  or  —3. 


3.  Given  a;* + 3a;' +  a;" --3a; =2  to  find  x. 

Ans.  «•=!,  —1,  —1,  or  —2. 

ip  Given  x* — Qx^  +  5x^  +  2a;' = 1 0  to  find  x. 

Ans.  x=:5,  —1,  or  1±  V^Tl, 


344  BIQUADRATIC   EQUATIONS. 

5*  Given  x*—4:X^—8x+S2=0  to  find  x. 

Ans,  x=2,  4,  or  —ldtV^» 

6.  Given  a;*— 9a;' +  30a;''— 46a? +  24=0  to  find  x. 

Ans.  x=l,  4,  or  2±/^. 

7.  Given  a;*— a;' + 2a;'' +  a; =3  to  find  x. 

Ans,  a;=±r,  or  ^^^'^^ 

8.  Given  6a;*— 43a;»  +  10'7a;»— 108a;  +  36=0  to  find  x. 

Ans.  a;=|,  li  2,  or  3. 

9.  Given  a?*+a;'— 16a;''  — 4a; +48=0  to  find  x. 

Ans.  a;=2,  3,  —2,  or  —4. 

10.  Given  a;*— a;'— lla;''  +  9a;+18=0  to  find  x. 

Ans.  xz=z2^  3,  —1,  or  —3. 

11.  Given  a;*— 8a;»  +  14a;'  +  4a;=8  tofinda;. 

Ans.  a;=3±V6,  or  litV3. 

12.  Given  a;*-12a;'  +  48a;'— 68a;+16=0tofind  x. 

Ans.  a?=3,  5,  or  2=fcV'3, 

13.  Given  2a;*— 2a;'— 2a;' H +  -=0  to  find  x. 

2      8 

1  ±i/2 
Ans.  a;-±i|/3,  or— -— . 

14.  Given  a;*  +  a;'— 29a;'— 9a;+180  =  0  to  find  x. 

Ans.  a;=3,  4,  —3,  or  —5. 

15.  Given  a;*— 4a;'— 29a;'  +  156a;=180  to  find  x. 

Ans.  a;=2,  3,  5,  or  —6. 

16.  Given  a;*  +  29a;'  +  287a;'  +  1147a;+1560=0  to  find  a;. 

Ans.  a;=  — 3,  —6,  —8,  or  —13. 

t<^    n-  4     n  3  .    ^Sa;'     27^;     81       ^    , 

17.  Given  a;*— 9a;'  +  — —  +-— =—  to  find  x. 

4  2         4 

Ans.  a;=li  li,  or  3  ±31^2. 

18.  Givena;*— 8a;'  +  23a;'— 64a;+120=0  tofinda:. 

Ans.  a;=3,  5,  or  ±2V^2. 

19.  Given  a;*— 16a;'  +  '79a;'— 140a;  +  68=0  to  find  x. 

Ans.  xz=2±V\  or  6  iV'Y, 


BIQUADRATIC   EQUATIONS.  345 

20.  Given  x*-5x'—5x^  +  45x=SQ  to  find  x. 

Ans.  x=l,  3,  4,  or  —3. 

21.  Given  a;*-3a;'— 15a;'  +  49a;=12  to  find  x. 

Ans.  x=S,  — 4,  or2±V3. 

22.  Given  x*-\-x^—x''—5x+4:—0  to  find  x.  

-3±1/-'7 
Ans.  ar=l,  1,  or . 

23.  Given  x* -{-x^—x*  + 1  Oa;  +  4 = 0  to  find  x. 

Ans.  X— ,  orlif"— 3. 

24.  Given  a;*  —  7a;'  +  Qx"  +  27a;=:54  to  find  x. 

Ans,  x=Z,  3,  3,  or  —2. 

25.  Givenar*  +  3ar'— 7a;'— 27a;=18tofindar. 

Ans.  a;=3,  —1,  —2,  or  —3. 

26.  Given  6a;*— 25a;' +  26a;' 4- 4a; =8  to  find  x, 

Ans.  a;=|,  2,  2,  or  -J. 

27.  Given  8a;*— 38a;»  +  49a;'— 22a;  +  3=0  to  find  x. 

Ans.  x=z\^  ^,1,  or  3. 

28.  Given  a;* — 9a;'  + 1 7a;'  +  27a;= 60  to  find  x. 

Ans.  a; =4,  6,  or  ±VS, 

29.  Given  a;*  +  a;' — 24a?'  +  43a; = 2 1  to  find  x.  _ 

^              ,    o        -5±|/53 
Ans.  a;=I,  3,  or , 

i 

30.  Given  a;* -fa;' 4- a;'  — 120a;  =100  to  find  a;.  

-5±5f/-5 

Ans.  a;=2±2V^2,  or . 

2 

3L  Given  x^ -{■  x^ -\- x" -\-  141a;= 100  to  find  x.    __ 

5  4-4^41  

Ans.  x= f-^— ,  or  2±4/-21. 

32.  Given  a;*-f-a;'  — 19a;'  — 49a;=30  to  find  x, 

Ans.  x=5,  —1,  —2,  or  —3. 

33.  Givena;*— a;'— 19a;'  +  49a;=30  to  find  ar. 

Ans.  a;=l,  2,  3,  or  —3. 

34.  Given  x*—^^x'-\-^x^—\^x  +  ^  —  0  to  find  x. 

Ans.  a;=|^,  i,  1,  or  3. 

35.  Given  a;*-38a;'  +  210a;'4-5.S8a;  +  289=0  to  find  x.         

Ans.  x=-l,  -l,or  20±V'lll. 


846  BIQUADRATIC  EQUATIONS. 

36.  Given4fl?*— 14a;'— 6a;»  +  31ar+6=0tofinda?. 

Ans.  x=2j  8,  or . 

4 

87.  Given  a;*— 6a;'~68a;'~114a;=ll  to  find  x.  

Ans.  x=^±^VS±j/l1±si^VS, 

(355*)  This  method  of  solution  is  applicable  to  all  afiected 
biquadratic  equations,  as  is  shown  by  the  solution  of  the  following 
literal  equation.  But  it  is  not  always  practicable,  as  the  quantity  to 
be  added  is  frequently  of  such  a  character  that  it  can  not  be  easily 
found. 

PROBLEM. 

Given  x*'-{a+h  +  c+d)x'  +  (ab  +  ac  +  ad-\-bc  +  bd-^cd)x^-'{abc-^ 
abd + acd  +  hcd)x + dbcd=0  to  find  x. 

SOLUTION. 
4a;*— 4 (a  +  6  +  c  +  d)x''  +  4(a6  +  ac -\- ad  +  bc-^  bd  +  cd)x^—  4  (abc  + 

abd  4-  cicd  +  bcd)x  +  4a6cc?= 0, 
[2x^—(a'{-b-{-c-\-d)xy—(a''—2ab—2ac—2ad  +  b^—2bc—2bd  + 

c' — 2cd + d')x'' —4:{abc  +  abd  +  acd  +  bcd)x  +  4a6cc? = 0, 
[2a;'— (a  +  6  +  c+c?)a;]'+  2{ab  +  cd) {2x^  —  {a  ■^b  +  c+d)x^  -f  a'i'  + 

2abcd-{-c^d''={a-\-b—c^dYx''—2{a  +  b^c—d)(ab—cd)x-\- 

a''b^—2abcd  +  c^d\     .-. 
2a;'—  (a+b-\-c+d)x  +  ab-\-cd  =  (a-{-b—c—d)x—  {ab—cd)j   or 

ab-'Cd—(a  +  b—c—d)x, 
23^—2{a  +  b)xz=—2ab;     or  2x^—2{c  +  d)x=-'2cd, 

a-\-b±(a—b)             ,             c-^d±(c—d)  _ 

x= -^ ^  =a,  or  0       X— -> ^—Cy  or  a. 

A  close  inspection  of  this  solution  will  enable  the  student  to  see 
what  relation  the  coefl5cient  of  a;',  in  the  added  square,  bears  to  the 
values  of  x,  as  finally  ascertained. 


MISCELLANEOUS    QUESTIONS.* 

1.  A  vintner  draws  a  certain  quantity  of  wine  out  of  a  full  vessel 
that  holds  256  gallons ;  and  then  filling  the  vessel  with  water,  draws 

*  These  questions  should  be  solved  without  resorting  to  the  method  just 
given. 


BIQUADEATIC  EQUATIONS.  847 

off  the  same  quantity  of  liquor  as  before,  and  so  on,  for  four  draughts, 
when  there  were  only  81  gallons  of  pure  wine  left.  How  much  wine 
did  he  draw  each  time  ?  Ans,  64,  48,  36,  and  2*7  gallons. 

2.  An  upholsterer  has  2  square  carpets  divided  into  square  yards 
by  the  lines  of  the  pattern.  Now,  he  observes  that  if  he  subtracts 
from  the  number  of  squares  in  the  smaller  carpet,  the  number  of 
yards  in  the  side  of  the  other,  the  square  of  the  remainder  will  exceed 
the  difference  of  the  number  of  squares  in  the  smaller  carpet,  and  the 
number  of  yards  in  its  side,  by  88.  Also,  the  difference  of  the  lengths 
of  the  sides  of  the  carpets  is  6  feet     What  is  the  size  of  each  carpet  ? 

Ans.  16  and  36  square  yards. 

3.  A  man,  playing  at  hazard,  won  at  the  first  throw  as  much 
money  as  he  had  in  his  pocket ;  at  the  second  throw  he  won  5  shil- 
lings more  than  the  square  root  of  what  he  then  had ;  at  the  third 
throT^  he  won  the  square  of  all  he  then  had,  and  then  he  had 
£112  I6s»    How  much  had  he  at  first  ? 

Ans.  18,  or  24^  shillings. 


CHAPTER  Xiy. 
HIGHER  EdUATIONS. 

(356*)  Equations  of  the  fifth  degree^  formerly  called  sur solid 
equations  and  equations  of  higher  degrees,  have  not  as  yet  been  found 
to  be  susceptible  of  any  general  solution.  Particular  examples,  how- 
ever, frequently  occur  that  may  be  reduced  by  known  methods. 
It  is  the  object  of  this  chapter  to  present  some  of  them. 

PROBLEM 

1.    Given  «'=«'  to  find  the  values  of  x, 

SOLFTION     I. 

(a:«+a«)(a:'-a')=0. 

(a?+a)(ar'— aa;  +  a") =0, 
a:'— aa:  +  a'=0, 

x=—- . 

x+a=zOj 
x=—a, 
x'-a'=0, 
{x—a){a^+ax-\-a')=0, 

x= . 

a?— a=0. 


x=a. 


a±aV—3         —a^aV—i 
xz=zay  —a, ,  or  - 


2 


HIGHEB   EQUATIONS.  849 

SOLUTION     II. 

<a?— a)(a?*+aa;*  +  aV  +  aV  +  a*x  +  a') =0, 

a;— a=:0, 

x=za, 

a?* + arc*  +  a V  4-  a'a;'  +  a*a;  +  a' = 0, 

(a? + a)x^ + a'x'{x  -\-a)-{-a*{x  +  a)= 0, 

(a?+aXa?*+aV  +  a*)=0, 

a;  +  a=0, 

a;=— a, 

aj*4-aV  +  a*=0,  

a;^ = . 

2 

Applying  the  rules  for  finding  the  square  root  of  surds,  we  get 

adzaV^S         — arhaV^^ 

x= ,  or . 

2         '  2 

It  is  easier,  however,  to  get  these  values  of  x  by  considering 

x'  +  a'x'  +  a*=(x'-^ay~a'x^ 

(a;»  +  a7-aV=0, 

(a;' +  a'— aa?)(a;'+a' +  aa;)=0, 

a?*— aa;+a'=0, 

and    a:*  +  aaj  +  a''=0, 


and 
2.  Given  a:*=:a'  to  find  x. 


2      

—a±aV—S 


2 

PROBLEM 
SOLUTION. 


a;=a 
(a:"-a)(a;*  +  eta;' +  aV +  a'a;  +  a*)=0. 
Placing  a;*4-aa?'  +  aV  +  a'a;  +  a*=0,  we  have  a  biquadratic  equa- 
tion in  which  the  coeflScients  are  literal.     To  solve  this  equation  re- 
quires an  artifice. 

Putting  x=ay,  and  we  have 

y'  +  -i  +  y-f— +  1=0,    puttin^y+-=2,and/4-— =2*— 2, 


860  HIGHER  EQUATIONS, 

we  have         2*— 2+0  +  1=0, 

1      — Idzl/'S 
2y»  +  (l=Fi/6)y=-2,  


l±t/5±i/-10:F2f6 


4   _ 
—adzaVt±aV- 

-10  +  2i^5 

i 
x=a. 

> 

—a  +  aVl+aV- 

-10— 2f6 

""                          4 

» 
-10+21^5 

""                           4 
— a+af5— at/- 

-10-21^6 

"^                           4  _ 
— a— af'S— af- 

» 
-10  +  21/6 

EXAMPLE  S- 

1.  Given  af'=l  to  find  a?.  

^Tis.  x=l,  -1,    ^ — -,  or . 

2*  Given  a?'=l  to  find  a;. 


-ldb|/5±V-10=F2*/6 

Ans.  a;=l,  or » 

4 

3*  Given  x*=.a*  to  find  x.  

Am,  xzzzdca,  ±aV—i,  ±aV~l,  or  ±aV—V'^.  . 

4*  Given  jr*=l  to  find  x. 

Ans.  x=±l,  ±V^,  ±V^,or  ±V-V'^. 
5,  Given  a!"+a'=:0  to  find  x.  

a±aV5±aV—10±2V5 

Ans.  «=— a,  or . 

4 

6t  Given  a;'  + 1  =0  to  find  x. 


^      .           ,         l±f^6±l/-10±2f^-5 
Ans.  a;=  —  1,  or . 


HIGHEB  EQUATIONS. 
7«  Given  aj*=a'  to  find  x. 


tn 


Atis. 


"  x=a,  ah,  ab*y  ah^,  ab^^  ah*, 
ah',  aV,  or  ah^,  in  which 


8.  Given  aj"=:a"  to  find  a;. 


Am, 


X=z- 


■a±aV6±aV—lQ^zV6 


a±aV5±aV-\0±2V6 


(357.)  The  following  examples  may  be  solved  by  a  combination 
of  the  principles  already  learned.  Some  of  them  are  inserted  for  the 
first  time  in  an  American  work,  and  will  be  found  to  be  the  most  diffi- 
cult algebraic  problems  that  have  ever  been  published  in  any  work  upon 
this  subject.  Many  of  them,  however,  will  be  found  to  be  easy  of 
solution.    Some  of  the  values  in  some  of  the  examples  are  omitted. 

PROS  LEM. 


Given  2i/y'''  +  6y— 2y''~-46y*— 6y +6'=0  to  find  four  values 
of  «. 


SOLFTIO 


2f^y"4-6y-2y'-46/-6y  +  6'=0, 

2yWfW-\-h'=2y'  +  4fty*  +  ^V, 
y'  +  2fVfTh'-¥f  +  h'=4.f  4-46/  +  6y, 
y'+V7Tl^=±{2f  +  hy) 
V'f  +  h'^f  +  hy, 
y'  +  b'=y'-\-2hy*  +  bY, 
2hy'  +  hY=h\ 
2y'  +  hf=h\ 

4y'=26,  or  — 46, 
2y=±i/26,  ±W^, 


(^), 


362  HIGHER  EQUATIONS. 

By  taking  the  minus  value  in  (A)  we  should  obtain  An  equation 
of  the  sixth  degree,  therefore,  the  given  equation  is  of  the  tenth  degree. 

EXAMPLES. 

1.  Given  2x^{x^-\-a^)^=2x\x-\-2a)-\-a\x—a)  to  &id  two  of  the 
five  values  of  a;.  Ans.  x=^aj  or  —a. 

2.  Given  a:'— 4a;' =621  to  find  all  the  values  of  a?. 

A              o        z,7^  --3±3f^II^        V23±l/2aK'^ 
Ans.  x=3y  — V23, ,  or ^ . 

3.  Given  a;"— 6a:'=16  to  find  all  the  values  of  x, 

A             o       s/o   -2=F2>/^       V2:FV2V/^ 
Ans.  x=2,  -  V2, ,  or ^ . 

4.  Given  x^  +x^  =756  to  find  all  the  values  of  x. 

Ans,  a;=243,  -28V'l8l,  243/""^'^         V  or  28Vl8lO^^-], 

5f  Given  a:'— a;2=66  to  find  all  the  values  of  a;. 

Ans.  a;=4,  V49,  1±V^,  or  V^izl^^-Y 

6*  Given  aa;8  +  6a;*  =c  to  find  two  values  of  x. 

Ans.  ^=±ME±^f. 
\  2a  / 

7.  Given  3a;"  +  42a;' =3321  to  find  all  the  values  of  a;. 

A              o       »/7T   -S±sy^       V41±'v/4U/^ 
Ans.  a;=3,  -'v/41, ,  or . 

40 

8*  Given  Vx^ ==3a;  to  find  all  the  values  of  a;. 


Ans.  a;=4,  V25,  -2 ±21^^,  or  V25(~^'^^~^^Y 
iren  (a;— 5)'— 3(a;— 5)2— 40  to  find  all  the  values  of  a;. 

Ans.  x=9,  5  +  V25,  3±2^/— 3,  or  54-V25[~^  ), 

17 
f  2=-—  to  find  all  the  values  of  x. 

^  

Ans.  a;=4,  '^I,  -2 ±2/^,  or  Vi(~^'^^^~^). 


8  17 

lOi  Given  -^-  +  2=-—  to  find  all  the  values  of  a; 
X  a;f 


HIGHER  EQUATIONS.  363 

IL  Given  x^-\ =— :=r+a;6  to  find  all  the  values  of  x, 

Ans.  a:=4,  1/49,  -lipV'^,  or  l/W"^^         ). 

a;*      ic'  1 

12«  Given  — —  =— — -  to  find  all  the  values  of  x. 

2        4  32 

Ans.  .=Vl  VI  Vi(=^^,  or  Vi(=^^). 
13t  Given  a;''  +  2'7a:'=2224  +  9a;*  to  find  all  the  values  of  a?. 

Ans..=  ±4,or±i/^l^^. 

14.  Given  (a;"  +  l)(a;'  +  l)(a;+l)=30a;'  to  find  all  the  values  of  a?. 

Am,  x=^~^,  or  -l±x/ir6±|/-liq:f^36 

15.  Given  (a;-i)»-— = ^^"^^  to  find  all  the  values 

^       '^       9      2(a?-i) +l/a:(a;-J)      _ 

„        ,   4±2f/l3  .  2  .-—- 

of  ar.  Ans,  a;=3,  —  i, ,  or  ±-v— 3. 


16.  Given  (l-a:)i/a/l+i]-2=V'a;+l+V'3a;-l  to  find  the  five 
values  of  a?.  Ans.  a;=l,  or ==:. 

ji 
A     5aj^ 
17i  Given  Zx^ =—592  to  find  the  eight  values  of  a;. 

2i 

Am.  x=±8,  ±8V^,  ±V^^^(W,  or  ±|/-V^=^(^. 

aj» _j   +(«" -\  =— to  find  the  eight  values  of  a?. 

^/is.  a;=±ar    — - — ,  or  ±aY  — - — » 


19*  Given  2a;^l— a;*=a(l+a;*)  to  find  the  values  of  a:. 


Am.  ar=±-|/-l±Vl-a*±|/2(lTV'l-a*. 


854  HIGHER  EQUATIONS. 

20.  Given  Sx^  +af«  =3104  to  find  the  ten  values  of  x. 


Ans.  a;=64,(Y)M6(  — 1±1/6±|/— 10qF2V^5,)  or 

21.  Given  ^ ^-f  ^ ^—=z —  to  find  x. 

a  X  c 

n 

acn+T 


Ans.  X: 


an+i    —  Cn+1 


22.  Given  "^   '='^'   '   to  find  x.         Ans.  x=(J^\  —"' 
n  8  \ms  ' 

28.  Given  ic^»-maf  =p  to  find  x.         Ans.  ^=\ra±Vm^  +  Ap'^i ^ 


24.  Given  a;"— 2aa:2=6  to  find  x.  Ans.  x={a±Va'-\-b)'^. 

25.  Given  3a;*'*~2a^=25  to  find  x.  Ans,  a:=(l±^?^j". 

—    4af 

26.  Given  So;"  Vo(^ ==4:  to  find  x. 

Ans.  a;=(8)T^,  or  {—^^)~^, 

27.  Givena;*"— 2ic"*  +  fl;"=6  tofinda;. 


.             i/l±1/13       \/l±V-1 
Ans.  x=y  -- g— ,  or  V  ^ • 

28.  Given  (X^—2x^  +  x=za  to  find  x. 

l±:V3zt2V4aTi 

Ans.  x= , 

2 

JL  3.     *"+"  1 

29.  Given  a'6'a?»—4(a6)2 a; 2 »»'»=(«— 6)'fl:w  to  find  «. 

.„.„{<ii«fi:f,.,(d!%iffif. 

30.  Given  «^v/^:±:J  •  ?5Z_(v^+ v^)  to  find  x. 

Of  -J"  0 

2W 


(a-±zh\p-q 


SIMULTAJSTEOUS  EQUATIONS. 


855 


SIMULTANEOUS    EQUATIONS. 


(358.)  Given 


PROBLEM. 


y         Vy 
a;  +  8=4y 


y 


to  find  the  values 
of  X  and  y. 


S  OLUTION. 


f       Vy    - 


iC'      2X        r-  ^ 


+Vy+i=±4i, 


-4-f/y=4,  or  —6, 

y 

x+yVy=^y,  or  -6y, 

a;=:4y— yfy,  or  — 6y— yfy, 
rr=4y— 8.       Second  eq.  transposed. 

4y— 8=4y— y^,  or  —5y—yVy, 


yVy=S; 
y*-8=0 
(yi-2)(y  +  23^i  +  4)=0 

y=4 

y+2yi=— 4 

yi=-l±V-S, 
y=—2zp2V^ 


0TyVy  +  9y=8, 

yf  +  9y=8, 

yt  +  l  =  -92/  +  9,     [yi  +  1. 

yf  + 1  =  — 9(2/— 1)  dividing 
y-yi  +  l  =  -9(2/i-l), 
y+8yi=8, 

yi=:-4±2f^6,  .V 

y=40±16/6, 


y=:-2(l=Fl/-3)But2/i  +  l=0, 


«       r.  X..  .•  .  i^=8,  -4,  -8(2±V^-3),orl52:F64»/6, 

By8ubstatutmg,weget  |^^^^      ^^  _2(ii v''^),or   40T16V^6. 


356 


SIMULTANEOUS  EQUATIONS. 


1*  Given  •> 


x'+y'= 


EX  A  M  P  LE  S. 

13    ^ 


xt/= 


X—1/ 

6 


x—y 


"  to  find  the  six  values  of  x  and  y. 


Ans, 


"  a;=3,  —2,  3a,  or  —2a, 
y=2,  —3,  2a,  or  —Za, 

m  wnicn  a= . 


2,  Given  |  ^^J'^^^H^'  |  to  find  the  six  values  of  x  and 


Ans. 


x=5,  4,  Sttf  or  4a, 
y=4,  5,  4a,  or  5ay 

in  which  a= '"'  , 


8»  Given 


a!'y--4=4ariy— |-, 


to  find  two  of  the  six  values 
js  111       I      f     of  X  and  y. 


fy2(a.2_y2)^ 


Am, 


=  1.  -1:f2/-2, 


ja:=l.  -1 
(  y=4,  -2 


4*  Given 


16aJ-y^=62^Ja^, 
a;*     12     a? 


to  find  the  values  of  x  and  y, 
L  y      ^'    V^'        J 

^w^    i^==±4,  ±16,  db2V^,  or  ±8|/^, 
•    (y=256,  (256)^  -192,  or  -3(64)'. 


r  2a;+y=26— 7t'2a;+y  +  4, 
5.  Given   -I  2^-H^_16     2x-y^  I   to  find  the  values  of  x 


[ 


2x—Vi/     A5     23;+^^^' 


—  lTi/321 
a:=2,  10, 5y ,  16,  —24,  or  ^  ^  , 


and  y. 

—  1^1^6145 


Ans. 


64 


64 


,    o^    161  ±1^321   ^,    ,,,        3073 ±^6143 
y=l,  26,  —32—,  64,  144,  or ^^ . 


6«  Given 


SIMULTANEOUS  EQUATIONS. 
^2t/^  —  SVx 


357 


Vx 


■\-V^f-16Vx=^Vx, 


Vx+V8{y-Vx)-4=y  +  l, 


to  find  the  values 
of  X  and  y. 


Ans. 


— 4T16I/-39                   788  ±24 1^644 
•^=4,  i g ,  A»  4,  or  — , 

„   ,     3±2V^=^     ,,        ,         87±|/644 
y=3,  If ,  If,  ~1,  or . 


7,  Given  \  ^1~^''X"^^V^^J      a     nn  !•    to    ^^^    ^"^^    sixteen 
values  of  x  and  y. 


^W5. 


a;=±3,  ±f6,  ±' 


,or 


/_-13±|/— 11 


2 

^      ,  l±V''^=l7  1±3^5       i±t/-Tr. 

y=2,  -1, ,  — ,  or 


2         '        2       '  2 

x-\-y  -{-xy  -\-iii?y  -\-xf  -\r  x^y  4-  2a;'y'  +  ajy"  +  «y 
+a:y=ll, 

8.  Given    ^  x'y  +  3a;y  +  3a;y  +  2a;y  +  4a;y  +  2a;y 

+  4:xY  +  4a;V*  +  xy*^ + a^y + «y  +  2a;y 
[  +a;y4-a:y=30 

to  find  the  sixteen  values  of  x  and  y.  

ix-\-y=^±\V^,\±\V~^,2orl,ov\  ±  V^, 
^"^^    -j      a:y=|:F^V/21,|:fi^-19,  lor2,orl  iFl^-2. 
Note. — ^The  solution  of  these  eight  simultaneous  equations  will  give  the  six* 
teen  values  of  x  and  y. 

9.  Given  x^yVxy—a^  xz^^xz—h^  y^zVzy=c  to  find  x^  y,  and  z. 

_  3     f  to  find  the  values  of  x  and  y. 
xy — c      \ 


I 


Ans. 


x=:(a^±Va"'-c^'')'^, 


y= 


(a"±  ♦/«""— c"')»* 


CHAPTER   XV. 
ARITHMETICAL  FBOGRESSION. 

(359.)  An  Arithmetical  Progression  is  a  series  of  quantities  in 
which  the  difference  of  the  consecutive  quantities  is  constant,  as 
-j-o  •  a±d  '  a±2d  •  a±Sd  •  a±4c?^  &c. 

PROBLEM. 

(360.)  To  find  a  general  expression  for  any  term  of  an  arith- 
metical progression, 

SOLUTION. 

\»U         2d,  3<^.  4<A,  Sth,  «<*, 

In  the  progression  a,  a±c?,  a±2a,  a±3a,  a±4a,  a±6a,  &c.,  we 
see  that  any  term  is  equal  to  a  plus^  or  minus  c?,  affected  by  a  coeffi- 
cient which  is  one  less  than  the  number  of  the  term ;  therefore,  if  we 
let  n  represent  the  number  of  any  term,  we  have  the  general  expres- 
sion 

nth  term  =za±.{n—\)d. 

If  we  suppose  the  progression  to  terminate,  we  have 
l—a±{n—\)d, 
in  which  /  represents  the  last  term,  n  the  whole  number  of  terms,  d 
the  common  difference,  and  a  the  first  term. 

PEOBLBM. 

(36 1 ,)  To  find  a  general  expression  fw  the  sum  of  all  the  terms 

of  an  arithmetical  progression. 

SOLUTION. 

Putting  S  equal  to  the  sum  of  all  the  terms  in  a  progresdon,  we 
have 

S=a-\-a±d-\-a±2d-}-a±3d /,  ion  terms,  (1). 

or,  S=l  +  lTd  +  l^:2d  +  l-:fSd a,     "       "       (2). 

2^=:(a  +  /)  +  (a+0  +  («  +  0  +  («  +  0 (^  +  «>    (3)=(l)  +  (2). 


ARITHMETICAL   PROGRESSION.  359 

Since,  2S  equals  (a  +  l)  taken  n  times,  the  expression  becomes 
2S={a  +  l)n, 

5=(»-±i)„or(a+0| 

which  is  the  expression  required. 

Remark. — By  the  aid  of  the  two  formulas   l=a±(n—l)dj  and 

S=l — ^  )w,  we  are  enabled  to  find  any  two  of  the  terms  a,  d,  n,  I,  3, 

when  three  of  them  are  given  since  we  shall  have  two  equations  and 
two  unknown  quantities.  We  append  a  few  simple  propositions  for  the 
student  to  demonstrate. 

PROPOSITION 

1.  In  an  arithmetical  progression  consisting  of  three  terms^  the 
sum  of  the  first  and  the  third  term  is  twice  the  second, 

PROPOSITION 

2.  In  an  arithmetical  progression  consisting  of  four  terms^  the  sum 
of  the  first  and  the  fourth  is  equal  to  the  sum  of  the  secoTid  and  the 
third, 

PROPOSITION 

8.  In  an  arithmetical  progression  consisting  of  any  number  of 
terms^  the  sum  of  any  two  terms  equally  distant  from  the  extremes  is 
equal  to  the  sum  of  the  extremes. 

PROPOSITION 

4.  In  an  arithmetical  progression  consisting  of  an  odd  number  of 
terms,  twice  the  middle  term  is  equal  to  the  sum  of  the  extremes. 

Remark. — ^In  the  following  examples  the  known  terms  should  be 
substituted  instead  of  the  letters  representing  them,  and  there  will  thus 
arise  one  or  two  equations,  according  as  one  or  both  of  the  formula, 

L—a±{n—  l)d  and  /S^=l  — —  k,  are  involved. 

QUESTION  s. 

1.  The  first  term  of  an  arithmetical  progression  is  2,  the  common 
difference  3,  and  the  number  of  terms  8.     What  is  the  last  term  ? 

Ans.  23. 

2.  The  first  term  of  an  arithmetical  progression  is  3,  the  common 
difference  2,  and  the  last  term  99.     What  is  the  number  of  terms  ? 

Ans.  49. 


860  ARITHMETICAL  PROGRESSION. 

3.  The  last  term  of  an  arithmetical  progression  is  100,  the  common 
difference  4,  and  the  number  of  terms  30.     What  is  the  first  term  ? 

Ans.  —16. 

4.  The  first  term  of  an  arithmetical  progression  is  —20,  the  num- 
ber of  terms  61,  and  the  last  term  230.  What  is  the  common  differ- 
ence ?  Ans.  5. 

5.  The  first  term  of  an  arithmetical  progression  is  —12,  the  com- 
mon difference  —7,  and  the  number  of  terms  101.  What  is  the  sum 
of  the  series?  Ans.  —36057. 

6.  Insert  8  arithmetical  means  between  3  and  21. 

Ans.  -T-5  •  7  •  9  •  11  •  13  •  15  •  17  •  19  . 

7.  Insert  3  arithmetical  means  between  i  and  |. 

Ans.  ^f.f^y.H. 

8.  The  sura  of  an  arithmetical  series  is  108,  the  first  term  3,  and 
the  common  difference  2.     What  is  the  number  of  terms  ? 

Ans.  10. 

9.  What  is  the  sum  of  w  terms  of  the  progression  -i- 1  •  2  •  3  •  4  •  <fec.  ? 

Ans.  S=l——\n. 

10.  The  first  term  of  an  arithmetical  progression  is  14,  and  the 
sum  of  eight  terms  is  28.     What  is  the  common  difference  ? 

Ans.  —3. 

11.  The  first  term  of  an  arithmetical  progression  is  12,  and  the 
common  difference  — i.  What  is  the  sum  of  the  series,  supposing  all 
its  terms  to  be  positive  ?  Ans.  150. 

12.  What  is  the  sum  of  the  series  -t-1-3-6-7'9*  to  100  terms  ? 

Ans.  10000. 

13.  The  first  term  of  an  arithmetical  progression  is  ^,  the  common 
difference  ^,  and  the  number  of  terms  25.  What  is  the  sum  of  the 
series  ?  Ans.  162i. 

14.  The  first  term  of  an  arithmetical  progression  is  1,  the  number 
of  terms  23,  and  the  sum  of  the  series  is  149^.  What  is  the  common 
difference?  Ans.  ^. 

15.  What  is  the  nth  term  of  the  series  -f-1  •  3  •  6  •  7  •  <fcc.  ? 

Ans.  271—1. 

16.  What  is  the  sum  of  n  terms  of  the  series  -j-l  *  3  •  5  •  7  •  &c.  ? 

Ans.  n*. 


AEITHMETICAL   PROGRESSION.  361 

17.  If  a  body  falling  to  the  earth  descends  a  feet  the  first  second, 
3a  the  second,  6a  the  third,  and  so  on.  How  far  will  it  fall  during 
the  ^th  second  ?  Ans.  {^t  —  \)a. 

18.  If  a  body  falling  to  the  earth  descends  a  feet  the  first  second, 
3  a  the  second,  5a  the  third,  and  so  on.  How  far  will  it  fall  in  t 
seconds  ?  Ans.  mf. 

19.  The  first  term  of  an  arithmetical  progression  is  -f,  the  common 
difference  is  1|,  and  the  number  of  terms  13.  What  is  the  sum  of 
the  series  ?  *  Ans.  139f . 

20.  The  first  term  of  an  arithmetical  progression  is  —J,  the  com- 
mon difference  —  f,  and  the  number  of  terms  25.  What  is  the  sum 
of  the  series?  ^ws.  —  281^. 

(36 2,)  There  are  problems  in  arithmetical  progression  to  which 
the  fundamental  formulae  are  not  immediately  applicable,  since  three 
of  the  quantities  a,  c?,  w,  ?,  s  are  not  given  to  find  the  other  two,  but, 
in  every  case,  the  number  of  terms  being  given  together  with  other 
conditions  to  find  the  terms. 

It  is  necessary  for  the  student  to  know  how  to  represent  in  the 
best  manner  a  series  of  numbers  in  arithmetical  progression.  One 
mode  has  already  been  given,  namely,  -^-x  '  x±iy  x±,2y'  x-^^y  (fee, 
in  which  x  represents  the  first  term  and  y  the  common  difference. 
This  mode  of  representation,  however,  is  seldom  the  most  expedient. 

When  the  number  of  terms  is  odd,  assume  the  middle  one  to  be 
equal  to  x,  and  y  the  common  difference  ;  thus, 
-^x^y*  X' x-\-y  '  when  there  are  three  terms. 

'T-x—2y'x—y'x'x-\-y'x-\-2y  "         "      five        " 

When  the  terms  are  even,  put  x—y  and  x-\-y  equal  to  the  middle 
terms,  2y  being  equal  to  the  common  difference  ;  thus, 
—«— 3y  •  x—y  '  x+yx  +  Sy  when  there  are  four  terms. 

-T-x—5y'x—Syx—y'x-{-y'x  +  Sy'x  +  5y'       "         "       six       " 

It  may  be  seen  that  in  this  mode  of  representation  the  common 
difference  disappears.  The  formula  for  the  sum  of  the  series  may  also 
be  easily  deduced  fi:om  this  method  of  representation. 

Thus,         -r- x—2yx—yx-x  +  yx-{-2y to  n  terms. 

and  thus,   -f- x—3y  •  x—y  •  x-\-y '  x-\-3y    to  n  terms. 

It  is  evident  that  in  either  of  these  cases  the  sum  of  the  series  is  n 
times  a;,  or  nx.    But  x  may  be  considered  equal  to  half  the  sum  of 

the  first  and  last  term,  or  = ,  therefore  S=n  x—{ \n. 


862  ARITHMETICAL  PROGRESSION. 


QUESTIO  N. 

What  four  numbers  are  there  in  arithmetical  progression,  of  which 
the  sum  of  the  squares  of  the  extremes  is  200,  and  the  sum  of  the 
squares  of  the  means  is  136  ? 

SOLUTION. 

Let  x  +  Sy,  x+y,  x—y^  x—^y  represent  the  numbers : 
then     2a;'' +  18^  =  200, 
and       2a;'  +  2/  =  136, 
16y'=64, 
42/=  ±8, 
y=-±.2, 
whence,  2a;'=:136— 2y''=:128, 
a:=:±8, 
.  •.  the  numbers  are  ±14,  ±10,  ±6,  and  ±2. 

QUESTIO  NS. 

1.  Four  numbers  are  in  arithmetical  progression.  The  sum  of  their 
squares  is  equal  to  2*76,  and  the  sum  of  the  numbers  themselves  is 
32.     What  are  the  numbers?  Ans»  11,  9,  7,  and  5. 

2.  A  number  consists  of  3  digits,  which  are  in  arithmetical  pro- 
gression ;  and  this  number  divided  by  the  sum  of  its  digits  is  equal 
to  26  ;  but  if  198  be  added  to  it,  the  digits  will  be  inverted.  What 
is  the  number  ?  Ans,  234. 

3.  The  sum  of  four  integral  numbers  in  arithmetical  progression  is 
20,  and  the  sum  of  their  reciprocals  is  |f .     What  are  the  numbers? 

Ans.  2,  4,  6,  and  8. 

4.  The  sum  of  $27  was  to  be  raised  by  subscription  by  three  per- 
sons, A,  B,  and  (7;  the  sums  to  be  subscribed  by  them  respectively 
forrainfr  an  arithmetical  progression.  But  C,  dying  before  the  money 
was  paid,  the  whole  fell  to  A  and  B ;  and  (7's  share  was  raised  between 
them  in  the  proportion  of  3:2,  when  it  appeared  that  the  whole  sum 
subscribed  by  A  was  to  the  whole  sum  subscribed  by  B::  4:  5. 
What  were  the  original  subscriptions  of  A,  B,  and  (7? 

Ans.  A's,  $3,  ^'s  |9,  and  C's  $16. 

6.  After  A^  who  went  at  the  rate  of  4  miles  an  hour,  had  traveled 
2f  hours,  B  set  out  to  overtake  him,  and  in  order  thereto  went  4^ 


I 


ARITHMETICAL  PROGRESSIOK.  863 

miles  the  first  hour,  4J  the  second,  6  the  third,  and  so  on,  gaining  ^ 
of  a  mile  every  hour.     In  how  many  hours  would  he  overtake  A  ? 

Arts.  8  hours. 

6.  The  base  of  a  right-angled  triangle  is  6,  and  the  sides  are  m 
arithmetical  progression.    What  are  the  other  two  sides  ? 

Am,  6,  8,  and  10,  or  4^,  6,  and  7^. 

7.  A  and  B  set  out  from  London  at  the  same  time,  to  go  round 
the  world  (23661  miles)  ;  one  going  east,  the  other  west.  A  goes  1 
mile  the  first  day,  2  the  second,  and  so  on.  B  goes  20  miles  a  day. 
In  how  many  days  will  they  meet,  and  how  many  miles  will  each 
have  traveled  ? 

Ans.  198  days.     A  goes  19701,  and  B  3960  miles. 

8.  A  traveler  sets  out  for  a  certain  place,  and  travels  1  mile  the 
first  day,  2  the  second,  and  so  on.  In  5  days  afterwards  another  sets 
out,  and  travels  12  miles  a  day*  How  long  must  the  second  travel  to 
overtake  the  first? 

Ans,  3,  or  10  days.     Explain  this  result* 

9.  A  and  B,  165  miles  distant  from  each  other,  set  out  with  a 
design  to  meet ;  A  travels  1  mile  the  first  day,  2  the  second,  3  the 
third,  and  so  on;  B  travels  20  miles  the  first  day,  18  the  second,  16 
the  third,  and  so  on.     How  soon  will  they  be  together  ? 

Ans.  In  10  or  33  days.     Explain  the  last  result. 

10.  There  are  four  numbers  in  arithmetical  progression  whose  con- 
tinued product  is  1680,  and  common  difierence  4.  What  are  the 
numbers?  Ans.  ±14,  ±10,  ±6,  and  ±2. 

11.  The  product  of  five  numbers  in  arithmetical  progression  is  945, 
and  their  sum  is  25.  What  are  the  numbers  ?    Ans.  9,  7,  5,  3,  and  1. 

12.  There  are  three  numbers  in  arithmetical  progression,  and  the 
square  of  the  first  added  to  the  product  of  the  other  two  is  16  ;  the 
square  of  the  second  added  to  the  product  of  the  other  two  is  14. 
What  are  the  numbers  ? 

Ans.  1,  3,  and  6  ;  or  —5,  —3,  and  —1. 

13.  There  are  two  casks,  A  and  B^  of  which,  A  the  greater,  holds 
312  gallons.  Into  A  a  certain  quantity  of  wine  is  put,  and  B  is  filled 
with  water ;  then  water  is  conveyed  out  of  B  into  A  in  the  following 
manner.  First,  a  number  of  gallons  is  taken,  which  is  less  by  2  than 
the  square  root  of  the  number  of  gallons  in  A  ;  then  a  quantity  less 


B64:  ARITHMETICAL  PROGRESSION. 

than  the  fonner  by  2  gallons,  and  so  on.  Now  when  B  is  in  this 
manner  exactly  emptied,  A  is  exactly  full ;  and  it  is  known  that  8 
gallons  were  taken  out  of  JB  at  one  time,  after  which  the  quantity  left 
in  JB  was  12  gallons.    What  is  the  number  of  gallons  of  wine  in  ^  ? 

Ans.  266. 

14.  From  two  towns  which  were  168  miles  distant,  two  persons,  A 
and  JB,  set  out  to  meet  each  other  ;  A  went  3  miles  the  first  day,  6 
the  next,  Y  the  third,  and  so  on ;  B  went  4  miles  the  first  day,  6  the 
next,  and  so  on.    In  how  many  days  did  th^  meet  ?  Ans*  8. 

15.  A  man  borrowed  $60  ;  what  sum  shall  he  pay  daily,  to  cancel 
the  debt,  principal  and  interest,  in  60  days;  interest  at  10  per  cent, 
for  12  months,  of  30  days  each  1  Ans,  $1  and  ^^  of  a  cent 


CHAPTER   XYI. 
GEOMETRICAL    PROGRESSION. 

(363.)  A  Geometrical  Proghiession  is  a  series  in  which  the 
successive  quantities  are  formed  by  multiplying  the  preceding  one  by 
a  constant  quantity,  which  is  called  the  ratio  of  the  progression  ;  as, 

-H-  2  :  4  :  8  :  16  :  32  :  64  in  which  the  ratio  is  2. 
-H-  27  :  9  :  3  : 1  :  i  :  I  in  which  the  ratio  is  i. 


Kemark. — ^The  ratio  of  a  geometrical  progression  is  the  constant  multiplier, 
and  it  would  be  more  philosophical  to  u^e  another  term,  as  the  French  do.  "We 
suggest  the  word  rate.  Some  EngHsh  writers  say  that  the  ratio  of  a  geometrical 
progression  is  the  inverse  ratio  of  its  consecutive  terms.  Briot,  a  French  writer, 
says,  that  "  TJie  (rapport),  ratio  of  each  term  to  the  preceding  is  called  the 
(raison),  rate."  a  few  say  that  the  direct  ratio  of  two  consecutive  numbers  is 
equal  to  the  second  divided  by  the  first.  This  is  not  only  unphilosophical  but 
is  not  consistent  with  the  symbol  used  to  express  ratio.  Thus,  a:b  =  c:dia 
read  the  ratio  of  a  to  6  equals  the  ratio  of  c  to  d.  Now,  the  symbol :  is  a  sign 
of  division,  and  is  generally  used  as  such  by  the  Germans  in  preference  to  the 
symbol  -i-  introduced  by  Dr.  Pell.  Several  American  writers  erroneously  call 
the  method  of  dividing  consequent  by  antecedent  to  express  the  ratio  of  the  latter 
to  the  former,  the  French  method.  Lacroix  is  the  only  French  author  that  we 
have  noticed  who  has  adopted  this  plan. 

PROBLEM. 

(364.)  To  find  an  expression  for  the  nth.  term  of  a  geometrical 

progression. 

SOLUTION. 

1st,       2^,      3<^.      4'*>       5'*»      6«A. 

In  the  series  -^^a  :  ar  :  ar^ :  ar^ :  ar* :  ar^ :  (fee,  it  may  be  seen  that 
any  term  is  equal  to  the  first  multiplied  by  the  ratio  afiected  by  an 
exponent  which  is  one  less  than  the  number  of  the  term. 
Therefore,  the  nth  term  =ar''~\ 


S66  GEOMETBICAL  PEOGRESSION. 

In  a  series  which  tenninates,  if  we  represent  the  number  of  terms 
by  n,  and  the  last  term  by  ^,  we  have 

"problem. 
(365.)  To  find  an  expression  for  the  sum  of  the  terms  of  a  geo- 
metrical progression. 

SOLUTION. 
Representing  the  sum  by  S^  we  have 

>S^=:a+ar+ar''-far'-far* ar'^'+ar^'-Hir'-',  (1). 

rS=ar-\ar^^r^^r^\ar^ ar"-'+ar"-'+ar*  (2)=(1)  x  r. 

then  rS-S=ar^—a  (3)=(2)-(l). 

Since  ar^=zar^^  xr,  and  ar'^^=l,  we  have  ar'^z^lr; 
.-.     rS—S=ar"—a, 
becomes  rS—S=lr—a 
(r-l)S=lr-a 

r—1 

which  is  the  expression  required.     When  r  is  less  than  1  it  is  best 

to  put  S= ,  although  the  same  result  will  be  obtained  from 

both  forms. 

PROB  LEM. 
(366.)  To  find  an  expression  for  the  sum  of  the  terms  of  a  de- 
creasing geometrical  progression  when  the  number  of  terms  is  infinite. 

SOLUTION. 
It  may  be  seen  from  the  formula  S= that    the   sum  of 

the  series  depends  upon  the  first  term,  last  term,  and  ratio. 
In  a  decreasing  geometrical  series  the  terms  must  continually 
approach  zero  as  a  limit.  Therefore,  when  the  number  of 
terms  is  infinite,  we  are  compelled  to  consider  zero  as  the  last 
term ;  since  there  is  no  quantity,  however  small,  greater  than 
zero  that  may  not  be  reached  or  passed  by  a  finite  number  of 
terms.  If,  then,  ^=0,  Ir  must  also  =  0,  and  the  above  for- 
mula becomes,  for  a  decreasing  geometrical  progression  having  an  in- 
finite number  of  terms,  S=- ,  which  is  the  basis  of  the  following 


GEOMETRICAL  PROGRESSION.  367 

RULE. 

Divide  the  first  term  of  an  infinitely  decreasing  geometrical  pro- 
gression by  the  difference  between  unity  and  the  ratio,  and  the  result 
will  be  the  sum  of  the  series. 

The  above  formula  may  also  be  deduced  in  the  following  manner : 
Let  S=ar\^r-\^r^+ar^,  &c.,  ad  infinitum,  r  being  less  than  1  (1) 

r;S^=  -f«r+a7-'-H»r^&c.,         "  "  "  "(2)=(l)xr 

rS-S=a  (3)  =  (2)-(l) 

S= -,  the  same  as  before. 

r—r 

A  few  simple  propositions  are  here  appended  for  the  student  to 

demonstrate. 

PROPOSITION 

(367.)  1.  In  a  geometrical  progression  consisting  of  three  terms 
the  product  of  the  extremes  is  equal  to  the  square  of  the  mean. 

PROPOSITION 

(368.)  2.  In  a  geometrical  progression  consisting  of  four  terms 
the  product  of  the  extremes  is  equal  to  the  product  of  the  means. 

PROPOSITION 

(369»)  3.  In  a  geometrical  progression  consisting  of  any  number 
of  terms  the  product  of  the  extremes  is  equal  to  the  product  of  any  two 
terms  equally  distant  from  them. 

EXAMPLES. 

1,  Find  the  11th  term  of  ^3  :  6  :  12  :  &c.  Ans.  3072. 

2.  Find  the  sum  of  9  terms  of  ^1  :  2  :  4  :  <fec.  Ans.  511. 

3t  Find  the  ratio  when  the  first  term  is  3,  last  term  768,  and 
number  of  terms  9.  Ans.  2. 

4,  Find  the  11th  term  of  4f  f  :  f  :  |-f  :  &«.  Ans.  tt f  ttt- 

5.  Find  three  geometric  means  between  4  and  324. 

Ans.  12,  36,  and  108. 

6,  Find  three  geometric  means  between  \  and  f . 

Ans.  ^|/6,  i,and|f^6. 

7.  Find  the  sum  of  -H-1  '.^:\'  &c.  to  infinity.  Ans.  2. 


368  GEOMETEICAL  PROGRESSION. 

8.  Find  the  value  of  .3333,  &c.,  or  j\  +  yf  o  +  i  o\o »  &c.,  to  infinity. 

Ans.  i. 

9.  Find  the  value  of  9.99999,  &c.,  or  9  +  A +  to  oj  <^c.,  to  infinity. 

Ans.  10. 

10.  Find  the  value  of  4f  J  :  —  f^  :  y^j  •  —/A  *  &c.  to  infinity. 

Ans.  -a. 

11.  Find  the  sum  of  -ff  1  :  -  :  —  ;  <fec.,  to  infinity.        Ans.    . 

X     x^  -^  '  x—l 

12.  Find  the  value  of  .2333,  &c.,  to  infinity.  Ans.  ^\. 

13.  Find  the  value  of  .3411111,  &c.,  to  infinity.  Ans.  f^J. 

14.  Find  the  value  of  .323232,  &c.,  to  infinity.  •    Ans.  ||. 

15.  Find  the  value  of  .20414141,  &c.,  to  infinity.  Ans.  f|ai. 

16.  Find  the  sum  of  the  series  40,  16,  &c.,  to  infinity. 

Ans.  66|. 

a'  x^ 

17.  Find  the  sum  of  -ffa;^  :ax:  —  :  &c.,  to  infinity.     Ans.  -w~~ . 

a  "•'^ 

18.  Find  the  sum  of  -tt-x^  : -j,  &c.,  to  infinity.     Ans. 


X"        '  "  icV-+a 

19.  Suppose  a  body  to  move  eternally  in  this  manner,  viz.,  20  miles 
the  first  minute,  19  miles  the  second  minute,  183^  the  third,  and  so 
on  in  geometrical  progression.  What  is  the  utmost  distance  it  can 
reach  ?  Ans.  400  miles. 

20.  What  is  the  distance  passed  through  by  a  ball,  before  it  comes 
to  rest,  which  falls  from  the  height  of  50  feet,  and  at  every  Ml  re- 
bounds half  the  distance  ?  *   Ans.  150. 

21.  In  the  preceding  problem,  supposing  that  a  body  falls  16|  feet 
the  first  second,  3  times  as  far  the  next  second,  and  5  times  as  far  the 
third  second,  and  so  on,  how  long  will  it  be  before  it  comes  to  rest  ? 

Ans.  j\%  1/579(4  +  34/2)  =  10-2t85222  seconds. 
^(3 70.)  There  are  many  interesting  problems  in  geometrical  pro- 
gression to  which  the  fundamental  formulae  do  not  immediately  apply. 
In  their  solution  a  great  deal  frequently  depends  upon  the  notation 
used. 

-^  x:x2/:  xy^  is  a  progression  of  three  terms,  y  being  the  ratio. 

vj-  X  \Vxy  :  y  is  a  progression  of  three  terms,  y?^  being  the  ratio. 


GEOMETRICAL   PEOGRESSION.  889 

This  last  method,  however,  may  be  best  explained  by  the  principle 
in  Prop.  1,  (367.) 

-fra; :  xy  :  xy^ :  xy^  is  a  progression  of  four  terms,  y  being  the  ratio. 

-ff —  \x:y  '.  —  is  a  progression  of  fom*  terms,  -  being  the  ratio. 

-7^  x'.xy  :  xy^ :  xy^ :  xy'^  is  a  progression  of  five  tei'ms,  y  being  the 
ratio. 

vr —  :  a;" :  a;y :  y^ :  —  is  a  progression  of  five  terms,  --  being  the  ratio. 

^  x\xy\  xy^ ;  xy^ :  xy^  :  xy^  is  a  progression  of  six  terms,  y  being 

the  ratio. 

,.  x^    x^  y^    y* .  y     ' 

-rr— ^  :  —  '.x'.y\  —  '  — ^  IS  a  progression  of  six  terms,  -  being  the 
y      y  X      X  X 

ratio. 

(371.)  It  is  sometimes  expedient  to  employ  substitution  in  the 
solution  of  geometrical  problems.     For  example,  if  we  put 

x+y=8, 
and     xy^='p^ 
we  get    x^  +  y' = s" — 2jo, 
and     ar'+y'^s'— 3^«, 
and    a;*  +  y* = s* — 4s'>  +  2p*, 
and     x^-^-y^zzz  s"  -^5s*p  +  5sp\ 

Remark. — ^It  would  be  a  good  exercise  for  the  student  to  ascertain  how  these 
results  are  obtained. 

QUESTION. 

(37  2»)  What  six  numbers  in  geometrical  progression  are  those 
of  which  the  sum  of  the  extremes  is  99  and  the  sum  of  the  other  four 
terms  90  ? 

SOLUTION. 

The  conditions  show  that  the  sum  of  the  six  numbers  is  189. 

Let  ar,  ary,  xy\  xy*,  xy*,  xy^,  represent  the  numbers.  * 

The  formula  S= —  becomes  by  substitution 

,««     xy'—x    x(y*^l) 
189=    ^    ,    =  ^       ,  S 
y—l  y—\ 

24 


870  GEOMETRICAL  PROGRESSION. 

But     xi/^-hx=99, 
99 


^■^  +  1' 
189(y--l)_    99 

21  11 

y*  +  f  +  l-y^^f+f-y  +  V 

21y*-21y'+21y''~21y  +  21  =  lly*  +  lly^  +  ll, 

10y  +  10y''  +  10=:21y^+21y  (a), 

10(y*  +  y'«  +  l)=21y(y'^  +  l), 
Putting  y'  + 1=0,  we  have  10(2'— y'')=21y2, 
IO2'— 21y2=10y^ 

21y±292/     6y 


20 


/  +  1='^ 


2' 

2y''-5y=-2, 

6±3     ^ 
y=— =2, 

_    99        99_ 

''-yTi^ss-^' 

Therefore,  the  progression  is  -H-3  :  6  :  12  :  24  :  48  :  96. 

Note. — Equation  (C)  is  recurring,  and  might  be  solved  according 
to  either  of  the  methods  given  in  biquadratics.  Equation  (C)  might 
have  been  obtained  without  using  the  general  formula  for  the  sum  of 
the  series. 

ANOTHER     SOLUTION. 

.8 


^       re'    ar'  y'    y'  ,  , 

Let  — 5,  — ,  Xy  y,  -^,  -^  represent  the  numbers. 

•••    4+^=99  (1), 

y'      x^  ^  '^ 

x^  v' 

and     — +  a;  +  y  +— =90  (2), 

y  ^ 

a;'+y^=99ary         (3)=(l)  x  xY,     Putting  a;  +  y=« 

and  xy=:py  we  have  s^ — Gs*/?  +  dsp'^=99p''  (4),  x  rcy 

x^  +  a:y  (ar  +  y)  +  y"  =  90a:y  (5)  ■=  (2), 

.T^  4-  y^ = 90iry — .ry(a:  +  y)      (6)  ==  (5)  transposed. 

«'— 35^=90/?— sp,  [forward 


GEOMETEICAL   PEOGEESSION.  871 


58'  5s         _      99s' 

i+2s"^(90+  2sY~(90  +  i 
5s  65'  995 


• 2^-  +  -^f——---^^l-—  (10)=(9)substitute(iin(4). 


90+  25  "^(90+  25)='"  (90  +25)'^' 
8100  +  3605  +  45''— 4505— 10s''+65*'=995, 
s'  + 1895=8100, 

— 189db261 


=36, 


5'  36-36-36      _  ,  ,      ,^^, 

p= = =18-4-4=12*24, 

^     90+25        2-9-9  ' 

a;  +  y=:36, 

iry=:  12-24.     Whence,  it  is  obvious  without 

solution,  that     a:=12, 

and     2/=24. 

Therefore,  the  series  is  — 3  :  6  :  12  :  24  :  48  :  96.   - 

Eemark. — ^This  problem  is  one  of  the  most  difficult  of  those  generally  pro- 
posed in  geometrical  progression,  and  the  solutions  given  should  be  carefully 
studied  by  the  student  that  he  may  be  able  to  solve  others  of  like  character. 

QUESTIONS. 

1.  The  sum  of  the  first  and  third  of  four  numbers  in  geometrical 
progression  is  148,  and  the  sum  of  the  second  and  fourth  is  888. 
What  are  the  numbers  ?  Ans.  4,  24,  144,  and  864. 

2.  There  are  three  numbers  in  geometrical  progression,  the  sum  of 
the  first  and  second  is  15,  and  the  dift'erences  of  the  second  and  third 
is  36.     What  are  the  numbers  ?  Ans.  3,  12,  and  48. 

3.  What  three  numbers  are  there  in  geometrical  progression  whose 
sum  is  14,  and  the  sum  of  whose  squares  is  84  ?    Ans.  2,  4,  and  8. 

4.  What  three  numbers  are  tho§e  in  geometrical  progression,  whose 
sum  is  52,  and  the  sum  of  whose  extremes  is  to  the  mean  as  10  to  3  ? 

Ans.  4,  12,  and  36. 

6.  What  three  numbers  are  those  in  geometrical  progression,  whose 
sum  is  13,  and  the  sum  of  whose  extremes  multiplied  by  the  me^n  is 
80  ?  Ans.  1,  3,  and  9. 

6.  The  sun;  of  the  first  and  second  of  four  numbers  in  geometrical 


372  GEOMETRICAL   PROGRESSION. 

progression  is  15,  and  the  sum  of  the  third  and  fourth  is  60.     What 
are  the  numbers  ? 

Ans.  5,  10,  20,  and  40  ;  or  —15,  30,  —60,  and  120. 

7.  The  sum  of  four  numbers  in  geometrical  progression  is  equal  to 
the  common  ratio  +1 ;  and  the  first  term  =xy*  What  are  the  num- 
bers ?  Ans.  y'y,  y\,  |f ,  uud  ^. 

8.  A  gentleman  divided  |210  among  three  servants;  the  sums 
received  were  in  geometrical  progression,  and  the  first  received  |90 
more  than  the  last.     How  many  dollars  did  each  receive  ? 

Ans.  1120,  160,  and  $30. 

9.  The  sum  of  three  numbers  in  geometrical  progression  is  35,  and 
the  mean  term  is  to  the  difierence  of  the  extremes  as  2  to  3.  What 
are  the  numbers  ?  Ans.  5,  10,  and  20. 

10.  There  are  three  numbers  in  geometrical  progression,  the 
greatest  of  which  exceeds  the  least  by  16.  Also,  the  difierence  of 
the  squares  of  the  greatest  and  the  least,  is  to  the  sura  of  the  squares  of 
all  the  three  numbers  as  5  :  7.     What  are  the  numbers  ? 

Ans.  5,  10,  and  20. 

11.  The  sum  of  three  numbers  in  geometrical  progression  is  13, 
and  the  product  of  the  mean  and  sum  of  the  extremes  is  30.  What 
are  the  numbers  ?  Ans.  1,  3,  and  9. 

12.  The  diagonals  of  4  squares  are  in  an  increasing  geometrical 
progression,  and  the  product  of  the  squares  of  the  diagonals  of  the 
extremes  is  to  the  product  of  the  diagonals  of  the  means  as  a  side  of 
the  third  is  to  the  square  root  of  the  common  ratio  divided  by  4^2. 
What  is  the  diagonal  of  the  third  square,  and  the  common  ratio,  sup- 
posing their  difierence  equal  to  45  ? 

Ans.  81  the  ratio,  and  36  the  diagonal  of  the  3d  square. 

13.  The  difference  between  the  first  and  second  of  four  numbers  in 
geometrical  progression  is  36,  and  the  difference  between  the  third 
and  fourth  is  4.     What  are  the  numbers  ?     Ans.  54,  18,  6,  and  2. 

14.  There  are  three  numbers  in  geomstrical  progression,  the  sum  of 
the  first  and  second  of  which  is  9,  and  the  sum  of  the  first  and  third 
is  15.     What  are  the  numbers  ?  Ans.  3,  6,  and  12. 

15.  There  are  three  numbers  in  geometrical  progression,  whose  sum 
is  14  ;  and  the  sum  of  the  first  and  second  is  to  the  sum  of  the 
second  and  third  as  1  to  2.     What  are  the  numbers  ? 

uins.  2,  4,  and  8. 


GEOMETRICAL    PROGRESSION.  SIB 

IG.  There  are  three  numbers  in  geometrical  progression,  whose  con- 
tinued product  is  64,  and  the  sum  of  their  cubes  is  584.  What  are 
the  numbers  ?  Ans.  2,  4,  and  8. 

17.  There  are  four  numbers  in  geometrical  progression,  the  second 
of  which  is  less  than  the  fourth  by  24  ;  and  the  sum  of  the  extremes 
is  to  the  sum  of  the  means  as  7  to  3.     What  are  the  numbers  ? 

Ans.  1,  3,  9,  and  21. 

18.  The  sum  of  $700  was  divided  among  four  persons,  whose 
shares  were  in  geometrical  progression ;  and  the  difference  between 
the  greatest  and  least  was  to  the  difference  between  the  means  as  37 
to  12.     What  were  their  respective  shares  ? 

Ans.  1108,  1144,  $192,  and  $256. 

19.  A  company  of  merchants  fitted  out  a  privateer,  each  merchant 
subscribing  $100.  The  captain  subscribed  nothing,  but  was  entitled 
to  a  $100  share,  at  the  end  of  every  certain  number  of  months.  In 
the  course  of  25  months  he  captured  three  prizes,  which  were  in 
geometrical  progression,  the  middle  term  being  i  the  cost  of  the 
equipment,  the  common  ratio  the  number  of  mouths  which  entitled  the 
captain  to  his  $100  share,  and  their  sum  $1375  more  than  the  cost  of 
the  equipmept.  After  deducting  $875  for  prize  money  to  the  crew, 
the  captain's  share  of  the  remainder  amounted  to  }  of  that  of  the 
company.  What  was  the  number  of  merchants,  and  the  captain's  pay  ? 

Ans.  25  merchants,  and  captain's  pay  $100  at  the  end  of  every  6 
months. 

20.  There  are  four  numbers  in  geometrical  progression,  the  differ- 
ence of  whose  means  is  3,  and  the  difference  of  whose  extremes  is 
10^.     What  are  the  numbers  ?  Ans.  1^,  3,  6,  and  12. 

21.  The  sum  of  three  numbers  in  geometrical  progression  is  31, 
and  the  sum  of  their  square  is  651.     What  are  the  numbers  ? 

Ans.  1,  5,  and  25. 

22.  The  sum  of  four  numbers  in  geometrical  progression  is  16,  and 
the  sum  of  their  squares  is  85.     What  are  the  numbei-s  ? 

Ans.  1,  2,  4,  and  8. 

23.  The  sum  of  five  numbers  in  geometrical  progression  is  31,  and 
the  sum  of  their  squares  is  341.     What  are  the  numbers  ? 

Ans.  1,  2,  4,  8,  and  16. 

24.  The  sum  of  six  numbers  in  geometrical  progression  is  94,  and 
the  sum  of  the  second  and  fifth  is  27.     Wliat  are  the  numbers  ? 

Ans.  1,  3,  6,  12,  24,  and  48. 


874  GEOMETRICAL   PROGRESSION. 

25.  The  sum  of  six  numbers  in  geometrical  progression  is  63,  and 
the  sum  of  the  means  is  12.     What  are  the  numbers  ? 

Ans.  1,  2,  4,  8,  16,  and  32. 

26.  The  sum  of  six  numbers  in  geometrical  progression  is  1365, 
and  the  sum  of  the  extremes  is  1025.     What  are  the  numbers? 

Ans.  1,  4,  16,  64,  266,  and  1024. 

27.  Whiit  six  numbers  are  those  in  geometrical  progression  whose 
sum  is  815,  and  the  sum  of  whose  extremes  is  165'^ 

Ans,  6,  10,  20,  40,  80,  and  160. 

28.  What  number  is  that  which  being  severally  added  to  3,  19, 
and  51,  shall  make  the  results  in  geometrical  progression  ? 

Ans.  13. 

29.  $120  are  divided  between  four  persons,  in  such  a  way,  that 
their  shares  may  be  in  arithmetical  progression ;  but  if  the  second 
and  third  had  received  $12  less  each,  and  the  fourth  |24  more,  the 
shares  would  have  been  in  geometrical  progression.  What  was  the 
share  of  each  ?  Ans.  |3,  $21,  $39,  and  $57  respectively. 

30.  The  sum  of  three  numbers  in  geometrical  progression  is  7,  and 
the  difference  of  whose  difference  is  1.    What  are  the  numbers? 

Ans.  1,  2,  and  4. 


CHAPTER  XVII. 
PROPORTION. 

(373.)  Proportion  is  an  equality  of  ratios. 

(37  4»)  If  the  ratio  of  a  to  6  is  equal  to  the  ratio  of  c  to  rf,  these 
four  terms  constitute  a  proportion  which  is  usually  written  a\h\:c\d, 
and  is  read  a  is  to  6  as  c  is  to  d.  Sometimes  the  sign  of  equality  is 
used  instead  of  the  four  dots,  as  a  :  6=c  re?,  which  may  be  read,  the 
ratio  of  a  to  6  is  equal  to  the  ratio  of  c  to  d. 

We  may  consider  the  symbol  :  as  an  abbreviation  of  the  sign  -f- ; 
whence,  we  infer   that  a\h:\c:d\s  only  another  mode  of  writing 

a     c 
a-7-6=c-i-rf,  which  is  the  same  as  r  =-.     This  shows  that  every  pro- 
portion is  essentially  an  equation. 

(375.)  The  four  quantities  of  a  proportion  are  called  its  terms. 

(376.)  The  first  and  the  fourth  term  are  called  the  extremes,  and 
the  second  and  the  third  term,  the  means. 

(377.)  The  first  two  terms  of  a  proportion  are  the  first  couplet, 
and  the  other  two,  the  second  couplet. 

(378.)  The  first  term  of  a  couplet  is  called  the  antecedent,  and 
the  second  term  the  consequent. 

(379.)  Three  quantities  are  in  proportion  when  the  ratio  of  the 
first  to  the  second  is  equal  to  the  ratio  of  the  second  to  the  third. 

(380.)  The  second  quantity  is  called  a  mean  proportional  be- 
tween the  other  two,  and  the  third  quantity  a  tki7'd  proportional  to 
the  other  two. 

Thus,  in  the  proportion  a:b::b:c,bis  the  mean  proportional,  and 
c  the  third  proportional. 

(381.)  The  equality  of  more  than  two  ratios  maybe  thus  written, 
a:h:ic:d::e:f::g:h,  &c.,  which  may  be  read  a  is  to  6  as  c  is  to 
d,  as  eis  to/,  as  ^  is  to  h,  &c. 

Remark. — The  student  should  observe  that  the  demonstrations  of  the  follow- 
ing propositions  in  regard  to  proportion  are  based  upon  the  fact  that  every 
proportion  is  essentially  an  equation. 


376  PROPORTION. 

PRO  POSITION 

(382.)  1.  In  every  proportion  the  product  of  the  extremes  is  equal 
to  -the  product  of  the  means. 

DEMONSTRATION. 

Let  a'.b'.:c:d  represent  any  proportion.  We  are  to  prove  that 
€td=bc.     This  proportion  expressed  as  an  equation  is 

-=-  '(1). 

b     d  ^  ^ 

ad=bc  (2)  =  (l)  xbd.  Q,E.D, 

Or, 

Put  a=rb,  then  by  the  nature  of  a  proportion  c=ird.  The  propor- 
tion will  then  stand rb:b::rd:d. 

Multiplying  the  extremes  together,  we  have  rbd. 
Multiplying  the  means  together,  we  have  rbd. 
These  results  are  identical,  therefore,  the  proposition  is  true. 
Remark. — ^This  proposition  furnishes  the  test  of  a  proportion. 

QUESTION. 

Are  2,  4,  3,  7,  in  proportion  ? 

SOLUTION. 

Multiplying  2  by  7,  we  get  14,  and  4  by  3,  we  get  12,  which  are 
not  equal,  therefore,  by  the  foregoing  proposition  they  are  not  in  pro- 
portion. 

QUESTIONS. 

1.  Are  3,  7,  8,  11  in  proportion? 

2.  Are  8,  16,  4,  2  in  proportion  ? 

3.  Are  2a?,  3a?,  4a?,  6x  in  proportion  ? 

4.  Are  i,  -J,  yV»  i  ^^  proportion  ?   • 

5.  Are  i,  i,  i,  jV  iii  proportion  ? 

Remark. — If  any  term  of  a  proportion  is  unknown,  put  it  equal  to  x,  and 
form  an  equation  by  placing  the  product  of  the  extremes  equal  to  the  product 
of  the  means,  and  then  solve  the  equation  to  obtain  the  value  of  x.  When  one 
of  the  means  is  unknown,  it  is  most  convenient  to  put  the  product  of  the  means 
equal  to  the  product  of  the  extremes. 

PROPOSITION 

(383.)  2.  When  three  numbers  are  in  proportion,  the  product  of 
the  extremes  is  equal  to  the  square  of  the  mean. 


PROPORTION.  877 

QUESTIONS. 

It  Are  3,  4,  5  in  proportion  ? 
2»  Are  3,  6,  12  in  proportion? 
3*  Are  x,  Vxt/,  y  in  proportion  ? 
4.  Are  ax^  ahx,  hx  in  proportion  ? 
5*  Are  oar,  xVab,  hx  in  proportion? 

PEG  POSITION 

(384i)  3.  When  the  product  of  two  quantities  is  equal  to  thepro^ 
duct  of  two  other  quantities,  the  four  quantities  may  he  expressed  in 
the  form  of  three  different  proportions, 

DEM  O  NSTR ATI O  N. 

Let  ad=zhc.  We  are  then  to  prove  that  all  of  the  following  pro- 
portions are  true, 

a',hi\c\d, 
a  \  c  \\  h  \  d, 
b:a::d:c, 
The  equation  ad=zhc  may  be  put  in  the  following  forms : 

a  c 
h^d' 
a  b 
~c^d^ 
h_d 
a~  c* 
and  these  three  equations  respectively  give 

a\h:\c:d, 
a\c\\h'.d, 
h:a::d:c,    Q,  E.  D. 

Corollary. — Since,  on  the  supposition  that  ad— he,  we  get  the 
proportions 

a\c  wh'.d 

h'.a  w  d\c, 

we  infer  that  these  proportions  are  also  true  on  the  supposition  that 

a\h  w  c'.d, 
because  this  proportion  gives 

ad^=hc. 
From  this  fact  we  obtain  the  two  following  propositions  : 


378 


PROPORTION. 


PROPOSITION 

•   (385.)  4.    When  four  quantities  are  in  proportion^  the  first  is  to 
the  third  as  the  second  is  to  the  fourth. 

If  a  :  6  : :  c  :  c?, 
then  a  :  c  : :  b  :  d.     • 
This  is  called  proportion  by  AltevTiation, 

PROPOSITION 

(386,)  5.   When  four  quantities  are  in  proportion^  the  second  is 
to  the  first  as  the  fourth  is  to  the  third. 
1£  a:b  ::  c:  d, 
then  b:a  ::  d  :c. 
This  is  called  proportion  by  Inversion. 

PROPOSITION 

(387.)  6.  When  a  couplet  is  common  to  two  proportions,  the  other 
two  couplets  will  constitute  a  proportion. 
If  a  :  6  : :  m  :  w, 
and  c:d  ::  miUj 
then  a:b  :  :  c  :  d. 
Let  the  student  prove  this. 

PROPOSITION 

(388.)  1.  If  a  :  6  : :  c  :  c?,  then  are  the  following  proportions  true : 
ma :  mb  ::mc:  md 


ma  '.mb'.'.   c 

:d. 

a',  b  ::mc 

'.md 

ma :  h  wmc 

'.d 

a:mb::  c 

'.md 

ma  '.mb'.'.nc 

'.nd 

ma  '.nb'.'.  mc 

'.nd 

a  ^    b  ^^  c 

.  ^ 

m'  m" m 

'  m 

a      b 
— :  — :;  c 

m    m 

:d 

m 

'  m 

m             m 

'.d 

b 

d 

m 


PKOPORTION.  879 

Let  the  student  prove  these  proportions  to  be  true  by  an  applica- 
tion of  Prop.  1,  (382.) 

PROPOSITION 

(389.)  8.  When  four  quantities  are  in  proportion,  the  sum  of 
the  first  and  second  is  to  the  second  as  the  sum  of  the  third  and  fourth 
is  to  the  fourth. 

DEMONSTRATION. 

Let  a',h\'.c  :d  (1).     We  are  to  prove  that 

u+h  :  h  :\c+d\  d 

ad=hc  (2)=(1)  by  Prop.  1,  (382.) 

ad-\-hd=hc-\-  bd  (3)  =  (2)  with  bd  added  to  both  members. 

(a  +  b)d=b{c  +  d)  (4)  =  (3)  factored. 

BjTro^,3,(3S4:)a  +  b:b::c  +  d:d.     Q.  K  D. 

This  and  the  derivative  proportion  in  the  following  proposition  is 
called  proportion  by  Composition, 

PROPOSITION 

(390.)  9.  When  four  quantities  are  in  proportion,  the  sum  of  the 
first  and  second  is  to  the  first  as  the  sum  of  the  third  and  fourth  is  to 
the  third. 

Let  the  student  prove  this. 

PROPOSITION 

(391.)  10.  When  four  quantities  are  in  proportion,  the  difference 
between  the  first  and  second  is  to  the  second  as  the  difference  between 
the  third  and  fourth  is  to  the  fourth. 

DEMON  STRATI  O  N. 

Let  a:b'.'.c  :  d  (1).     We  are  to  prove  that 

a — b  '.b  '.'.  c — d  '.  d 

ad=:bc  (2)  =  (1)  by  Prop.  1,  (382.) 

ad—bd=bd—bc       (3)  =  (2)  with  bd  subtracted  from  both  members. 
{a—b)d=:b{c—d)      (4)  =  (3)  factored. 
By  Prop.  3,  (384)  a—b  :b::  c—d  :  d.      Q.  E.  D. 
This  and  the  derivative  proportion  in  the  following  proposition  is 
called  proportion  by  Division, 


880  PEOPORTION. 


PROPOSITION 


(392.)  11.  When  four  quantities  are  in  proportion^  the  difference 
between  the  first  and  second  is  to  the  first  as  the  difference  between  the 
third  and  fourth  is  to  the  third. 

Let  the  student  prove  this. 

PROPOSITION 

(393.)  12.  When  four  quantities  are  in  proportion,  the  sum  of 
the  first  and  second  is  to  their  difference  as  the  sum  of  the  third  and 
fourth  is  to  their  difference, 

DEMONSTRATION, 

Let  a:b  ::  c:d.    We  are  to  prove  that 

a  +  b  :a—b  : :  c-\-d  :  c—d^ 
By  Prop.  8,  (389)  and  Alternation,  a-\-b: c-{-d  ::  b  :  d 
By  Prop.  9,  (390)     «  "  a-b:c-d::b:d 

then  by  Prop.  6,  (387)     "  "  a  +  b:a—b::c+d:c—d 

Q,  E.  D. 

PR  OPOSITION 

(394.)  13.  In  a  continued  proportion,  any  antecedent  is  to  its 
consequent  as  the  sum  of  all  the  antecedents  is  to  the  sum  of  all  the 
consequents, 

DEMONSTRATION. 

Let  a  :b  ::  c  :  d  ::  e  :f::  ff  :  h  ::  &c.    We  are  to  prove  that 
a:  b::a-\-c+e-\-ffj  <fec. :  b+d-^-f+h,  &c. 

ad=:bc 

af=be 

ah=:bg 

&c.=&c. 

Whence,  ad-\-af-\-ah,  &c.=6c  +  6e+6^,  <fec 
Adding  ab,  we  have  ab-\-ad  +  af-\-ah,  &;c.=ab  +  bc-}-be-{-bg,  <feo. 

a(6+c?  +  /+^,  &c.)=b{a  +  c-he+g,  &c.) 
By  Prop.  3.  (384)a:6  ::  a4-c+e+^,  &c.  :  b+d-^-f+h,  &c 


By  Prop.  1,  (382) 


Q.  B.  D. 

PROPOSITION 

(395.)  14.  If  the  corresponding  terms  in  two  proportions  be  mul- 
tiplied together  the  products  will  constitute  a  proportion. 


PEOPORTION.  S81 

DEMONSTRATION. 

Let  a  :  b  : :  c  :  d^ 
and  m:n::p:q. 
We  are  to  prove  that  am  ibniicji):  dq. 

By  Prop.  1,(382.)    i  '^=^'' 
•^         j^     '  V  /J  mq=np, 

Multiplying  am'dq=:.hn'cp. 

By  Prop.  3,  (384.)  amibn: :  cp  :  dq.     Q.  K  t), 

PROPOSITION 

(396.)  15.  If  the  terms  of  one  proportion  be  divided  hy  the  cor- 
responding terms  of  another  proportion^  the  quotients  will  constitute 
a  pi'oportion. 

Let  the  student  prove  this. 

PROPOSITION 

(397.)  16.  If  four  quantities  be  in  proportion^  their  like  powers 
or  roots  will  be  in  proportion. 

DEMONSTRATION. 

Let  a  :  b  : :  c  :  d. 

(  a"* :  i*" : :  c** :  rf"*, 
We  are  to  prove  that  •(    i     i       i      i  * 

/  am  '.  bm  I  I  C«*  I  dm. 

By  Prop.  1,  (382.)  ad=bc, 

By  involution  a'"c?"*  =  6"*c*", 

L  L       L  L 
By  evolution  a»»<?'»=6mc»». 

Whence  by  Prop.  3,  (384.)  )    ji     i      i     ^  [      Q.  K  D. 


PROP  O  8  ITIO  N 

(398.)  Vl.  If  three  quantities  are  in  proportion^  the  first  is  to  the 
third  in  the  duplicate  ratio  of  the  first  to  the  second^  that  is  as  the 
square  of  the  first  is  to  the  square  of  the  second. 

K  a ;  6  : :  6  :  c,  prove  that  a :  c : :  a^ :  b^. 

PROPOSITION 

(399.)  1 8.  If  four  quantities  are  in  continued  proportimi^  the  first 
is  to  the  fourth  in  the  triplicate  ratio  of  the  first  to  the  second. 
l£a:b::b  :c::c*.d,  prove  that  a*.d::a^:b^. 


382  SAEMONICAL  PROPORTION. 

PROPOSITION 

(400.)  I9.1f  mm: :p:q  and  am:bn::c:d,  then  ap:bq::c:d. 
If  m:n::p:q  and  a:b::mc:  ndj  then  a  :  b  ::pc  :  qd. 
If  m:n  ::p:q  and  am  :b::       :d,  then  apib  : :  cq  :  d, 
Jfm:n::p:q  and  a:bm::c:  dn,  then  a:bp\',c\dq. 

PROPOSITION 

(401.)  20.  If  j";^;;^'^[  thena:6±e::c:t/±/. 

If  i«-f*-*^-^[  thena±e:6::c±/:e?. 
(  e  :  0  : :/ :  a  )    . 

PROPOSITION 

(402.)  21.  i)^  the  two  consequents  of  four  quantities  in  propoT" 
tion  be  increased  or  diminished  by  quantities  which  have  the  same 
ratio  as  the  antecedents^  the  resulting  quantities  and  the  antecedents 
will  be  in  proportion. 

If  i     •    •  •    •     (.  then  a  :  c  : :  6=tm  :  c^dcw. 
(  axcwmin  ) 


HAEMONICAL    PROPORTION. 

(403.)  Three  quantities  are  in  harmonical proportion^  when  the 
first  is  to  the  third,  as  the  diflference  between  the  first  and  the  second 
is  to  the  difierence  between  the  second  and  third. 

The  quantities  a,  6,  and  c  are  in  harmonical  proportion  when 
a\  c  '.  a^^b  :  b'^c. 

(404.)  Four  quantities  are  in  harmonical  proportion^  when  the 
first  is  to  the  fourth,  as  the  difference  between  the  first  and  second  is 
to  the  difference  between  the  third  and  fourth. 

The  quantities  a,  6,  c,  and  d  are  in  harmonical  proportion  when 
a\d\'.  a^^b  :  c^d, 

PROBLEM. 

(405.)  To  find  an  harmonical  mean  between  a  and  c. 

SOLUTION. 

Let  X  =  the  required  harmonical  mean.  Since  a,  x,  and  c  are  in 
Jiarmonical  proportion,  we  have  a:c::a—x'.x—c. 


HABMONICAL   PROGRESSION.  388 

ax — ac=zac — car, 

x=z ,  tiie  mean  to  be  found. 

PROBLEM 

1.  Given  the  first  and  second  of  three  quantities  in  harmonical  pro- 
portion to  find  the  third. 

PROBLEM 

2.  Given  the  second  and  third  of  three  quantities  in  harmonical 
proportion  to  find  the  first. 

PROBLEM 

3.  Given  the  first  three  of  four  quantities  in  harmonical  proportion 
to  find  the  fourth* 

PROBLEM 

4.  Given  the  last  three  of  four  quantities  in  harmonical  proportion 
to  find  the  first. 

PROBLEM 

6.  Given  the  first  and  last  and  one  of  the  middle  quantities  of  four 
quantities  in  harmonical  proportion  to  find  the  other  middle  quantity. 


^  >  t  »i » ,  »h 


HARMONICAL    PROGRESSION. 

(406.)  An  Harmonical  Progression  is  a  series  of  quantities, 
any  consecutive  three  of  which  are  in  harmonical  proportion. 

PROPOSITION 

(407  •)  1.  The  reciprocals  of  a  series  of  quantities  in  harmonical 
progression  are  in  arithmetical  progression. 

DEMONSTRATION. 

Let  a,  6,  c,  d,  e,f  &c.,  be  an  harmonical  progression.     We  are  to 

prove  that  -,  r,  -,  3,  -,  -;,,  &c.  is  an  arithmetical  progression. 
^  a   0  c   a    e  f 


384 


HABMONICAL  PROGRESSION. 


If  we  prove  that    * 


1     1_2 

1     1_2 
1     1_2 


1     1 


2 


(fee,  &c., 

we  shall  establish  the  truth  of  the  proposition. 

,f^   ,         ,       2ac  b       ac  2      a  +  c     1     1 

We  have  b=- — ,  or  -= — --,  or  -= ==-^ — - 

a+G        2      a+c         b        ac       c    a 


We  nave  b= ,  which  gives  -  +  -=--, 


c= 


26c? 


d= 


b+d' 
2ce 


1  1_2 
b~^d~~c' 
1     1_2 


6= 


1     1_2 


C.^.i). 


We  are  to  prove  that 


d-\-f 

ANOTHER     DEMONSTRATION. 
1_1_1_1 

a  b~  b  c' 
1_1_1_1 
6  c~  c  d* 
1111 


By  the  nature  of  the 
progression,  we  have 


Whence,  we  get 


c  d  d  e 
1__1_1_1 
d  e'~  e  f* 
&c.,      &c. 

a\c'.  \a—b  :  6— c, 
b'.d'.',  b—c  :c—dy 
c:e  : : c — d: d — e, 
d:f:  :d—e  :e—f, 

ab-^ac=ac—bcj 
bc—bd=:bd—cd, 
cd—ce  =  ce—de, 
de—df^df—efj 


HARMONIOAL   PROGRESSION". 


885 


Dividing  respectively  by    < 


def,  gives 


These  equations  by  transposition, 

(1     1  __!__! 
a     6~"  b     c' 
1_1_1     1 
b     e~~  c     (T 

c~~d'~d     i 

1-1=1-1.  <i.KD. 
d     e     e    J 


abc,  gives  --1=1--, 

1.  1      '        1111 
bed,  gives  j-^--y 

,       .        1111 

cde.  ffives ;=-: — , 

*  ^         e     d     d     € 

1111 


f     e"  e    dC 


become    < 


The  converse  of  the  preceding  proposition  is  evidently  true,  there- 
fore, we  have 


PROP  O  SITI  ON 

(408 •)  2.  The  reciprocals  of  a  series  of  quantities  in  arithmeti' 
cal  progression,  will  constitute  an  harmonical  progression. 

Since  1,  2,  3,  4,  6,  6,  <fec.,  is  an  arithmetical  series,  their  reciprocals 

T'  2'  3'  4'  5'  6'  ^^*'  ^  ^^  iiarmonical  series. 
The  fractions  in  this  series  are  in  the  ratio  of 

60,  30,  20,  15,  12,  10,  <fec., 
which  must  also,  be  an  harmonic  series. 

Remark. — It  is  a  principle  in  music  that  the  longer  a  string  is,  the  lower  is 
the  sound  produced  by  its  vibration.  If  then,  we  have  musical  strings  of  equal 
weight  and  tension  which  are  in  the  ratio  of  60,  30,  20,  15,  12,  10,  and  caU  the 
Bound  produced  by  the  vibrations  of  the  string  whose  length  is  60,  the  key  note, 
the  sounds  produced  by  the  stiing  whose  length  is  30,  will  be  the  octave  of 
this  key-note ;  by  the  string  20  the  fifth  of  this  octave,  or  the  twelfth  of  the 
key-note :  by  the  string  15,  the  octave  of  the  octave,  or  the  double  octave  of 
the  key-note ;  by  the  string  12  the  third  of  this  double  octave  or  the  seventeenth 
of  the  key-note ;  by  the  string  1 0  the  fifth  of  this  double  octave  or  the  nine- 
teenth of  the  key-note.  The  simultaneous  vibration  of  these  will  produce 
what  is  caUed  harmony ;  hence,  the  name  harmonical  progression. 

25 


386  HAEMOKICAL  PKOGRESSION* 

PEOBLEM. 

(409.)  To  find  any  number  of  harmonic  means  between  two 
quantities. 

SOL  UTION. 

Let  it  be  required  to  find  m  harmonic  means  between  a  and  c. 

By  Prop.  1,  (407.),  we  have  - -,  an  arithmetical 

progression  of  m  +  2  terms. 

By  substituting  in  the  formula  /'=adb(^— l)c?;  we  have 


Having  found  the  common  difierence,  we  are  now  able  to  insert  m 
arithmetical  means  in  the  arithmetical  series. 
1  1 

h c* 

The  series  becomes 

_^1    1  h—c         1         2(6--c)       1         3(6— c) 

'^  h'h    "^db(m+T)^*6"^  ±(m  +  l)6c*  i"^  ±(m  +  l)6c* 
1  4(6— c)       _    1 

6"^  dr(m  +  l)6c  ***? 
1      mc-^h      7nc-{-2b—c  wc 4-36— 2c   mc  +  46— 3c         1 
'6'(m  +  l)6c'    {m-\-l)bc  '     {m-{-l)bc   '    (m+l)6c  c 

Hence,  the  harmonic  series  is 

(m+l)6c     (m  +  l)6c  (m  +  l)6c  (m  +  l)bc  I 

'     mc  +  f'  (m— l)c+T6*   (m— 2)c  +  36'   (m— 3)c4-46        c' 

E  X  AMPLES. 

It  Find  an  harmonic  mean  between  3  and  6  Arts.  4, 


2.  Find  an  harmonic  mean  between  x+y  and  a:— y.      Ans, 


X 


3«  Find  an  harmonic  mean  between  — ; —  and .       Ans,  -, 

x-\-y         x—y  X 

4.  Find  the  third  of  three  quantities  in  harmonical  proportion,  the 

first  and  second  being  3  and  4.  Ans,  6. 


HAEMONICAL  PKOGRESSION.  887 

5*  Find  the  first  of  three  quantities  in  harmonical  proportion,  the 
second  and  third  being  144  and  104.  Ans.  234. 

6«  Find  the  fourth  of  four  quantities  in  harmonic  proportion,  the 
first  three  being  2,  3,  and  8.  Ans.  16. 

7.  Find  the  third  of  four  quantities  in  harmonical  proportion,  the 
first  being  10,  the  second  12,  and  the  fourth  15.  Ans.  12. 

8.  Find  the  second  of  four  quantities  in  harmonical  proportion,  the 
first  being  6,  the  third  9,  and  the  fourth  15.  Ans.  7. 

9.  Find  the  first  of  four  quantities  in  harmonical  proportion,  the 
second  being  6,  the  third  9,  and  the  fourth  15.  Ans.  4f. 

10.  Find  a  harmonic  mean  between  50  and  100.  Ans.  66|, 

11.  Find  a  harmonic  mean  between  25  and  50.  Ans.  33j. 

12.  Find  a  harmonic  mean  between  12^  and  25.  Ans,  16|. 

13.  Find  two  harmonic  means  between  1|  and  3.  Ans.  Ij^  and  2. 

II,  Find  three  harmonic  means  between  10  and  30. 

Ans.  12,  15,  and  20. 

15.  Find  three  harmonic  means  between  315  and  35. 

Ans.  105,  63,  and  45. 

16.  Find  fourteen  harmonic  means  between  |  and  4. 

^^«.  A,  f,  tV»  h  i*r,  h  h  h  h  h  h  1.  H.  and  2. 

17.  Find  the  fifth  term  of  an  harmonical  progression  whose  first  term 
is  60  and  second  term  21.  Ans.  '7j\. 

18.  Find  the  unknown  terms  of  an  harmonical  progression  consist- 
ing of  12  terms,  the  first  being  4  and  the  fourth  1. 

^^'  2,  11,  4,  I,  4,  1,  J,  I,  ^V,  and  J. 

19.  Find  the  resulting  proportion  when  a,  6,  c  are  in  arithmetical  pro- 
gression, and  6,  c,  d  in  harmonic  proportion.       Ans.  a  :  b  ::  c  :  d. 

20.  Find  the  wth  term  of  an  harmonical  progression,  a  and  b  being 
the  first  two  terms. 

ab 


Ans. 


«(%— 1)  — 6(w— 2)* 


888  PROBLEMS  IN   PROPOBTIOK. 

FBOBLBMS    IN    PROPORTION. 

PROBLEM. 

(41 O*)   There  are  two  numbers  whose  product  is  24,  and  the 

difference  of  their  cubes  is  to  the  cube  of  their  difference  as  19  is  to  1. 
What  are  the  numbers  ? 

80LUTI o  N. 


Let  a;  =  the  greater  numbw, 

and  y  =  the  lesser  number 

'. 

By  lat  condition,    x^—y^  :  {x—yf  : :  19  :  1 

x''+xy+y^  :  x'-'lxy^f  : :  19  :  1 

Prop.  7. 

(9) 

3a;y  :a;'+  a;y+y'::18  :  19 

Prop.  11. 

xy\^^-  xy  +  y-'w  6   :  19 

Prop.  1. 

(11) 

xyix'  +  'ixy  +  y'w  6   :  25 

Prop.  9  and 

Prop.  6. 

^xy:x''-\-2xy-\-y^::24:  :  26 

Prop.  1. 

(4) 

{x+yY:(x--yy::25:l 

Prop.  10  and  Prop.  6. 

x+y  :  x--y::  5  :  1 

Prop.  16. 

2x:2y::  6  :  4 

Prop.  12. 

X  :  y::  3  :  2 

Prop.  7. 

(8) 

8y-=2a: 

Prop.  1. 

y=%x 

By  2d  condition,                         a;y=24 

|a;'=24 

ic'=36 

x=±6 

y=±4 

QUESTION  S. 

1.  There  are  two  numbers  whose  product  is  135,  and  the  differ- 
ence of  their  squares  is  to  the  square  of  their  difference  as  4  to  1. 
What  are  the  numbers  ?  Ans.  15  and  9. 

2.  There  are  two  numbers  which  are  to  each  other  in  the  duplicate 
ratio  of  4  to  3,  and  24  is  a  mean  proportional  between  them.  What 
are  the  numbers?  Ans.  32  and  18. 

3.  There  are  two  numbers  whose  sum  is  24,  and  their  product  is  to 
the  sum  of  their  squares  as  3  to  10.    What  are  the  numbers  ? 

Ans,  18  and  6. 


PROBLEMS   IX   Pr.OPORTION.  889 

4,  There  are  three  numbers  which  are  to  each  other  as  3  to  2.  If 
6  be  added  to  the  greater  and  subtiacted  from  the  lesser,  the  smn  will 
be  to  the  remainder  as  3  to  1.    What  are  the  numbers  ? 

Ans.  24  and  16. 

6.  There  are  two  numbers  whose  sum  is  60,  and  their  product  is  to 
the  sum  of  their  squares  as  2  to  5.      What  are  the  numbers  ? 

Ans,  40  and  20. 

6.  The  number  20  is  divided  into  two  parts,  which  are  to  each 
other  in  the  duplicate  ratio  of  3  to  1.  What  is  the  mean  proportional 
between  these  parts  ?  Ana*  6* 

Y.  If  — 7 — =4a,  show  that  a-\-x :  2a : :  26  :  a—x. 


8.  If  (a+xy :  (a—xy  ::x-^i/:  x—y^  show  that  a\x\  :V2a—y :  Vy, 


9.  If  ar :  y : :  a' :  i"*  and  a:h\\\/c  +  xi  \/d-\-y  show  that  dx=cy, 

10.  l£x^  :y^::S6:  25  and  2X'\-y:x-\-2  in  a,  ratio  compounded  of 
the  ratios  of  1*7  :  2  and  2  :  7,  what  are  the  values  of  x  and  y  1 

Ans.  a?=12,  and  y=10. 

11.  A  person  has  British  wine  at  5  s.  per  gallon,  with  which  he 
wishes  to  mix  spirits  at  lis.  per  gallon,  in  such  proportion  that  by 
selling  the  mixture  at  9s.  a  gallon  he  may  gain  35  per  cent.  What 
is  the  necessary  proportion  ? 

Ans.  13  gallons  of  wine  to  5  of  spirits. 

12.  What  number  is  that  to  which  if  3,  8,  and  lY  be  severally 
added,  the  first  sum  shall  be  to  the  second  as  the  second  to  the  third  ? 

An^.  3i. 

13.  A  merchant  having  mixed  a  certain  number  of  gallons  of  brandy 
and  water,  found  that  if  he  had  mixed  6  gallons  more  of  each  there 
would  have  been  Y  gallons  of  brandy  to  every  6  gallons  of  water,  but  if 
he  had  mixed  6  gallons  less  of  eajch  there  would  have  been  6  gallons  of 
brandy  to  every  5  gallons  of  water.     How  much  of  each  did  he  mix  ? 

Ans.  is  gallons  of  brandy  and  66  of  water. 

14.  ^  and  B  speculate  in  trade  with  different  sums.  A  gains 
1150,  B  loses  |50,  and  now  ^'s  stock  is  to  ^'s  as  3  to  2  ;  but,  had 
A  lost  $50  and  B  gained  $100,  then  ^'s  stock  would  have  been  to 
J^'s  as  5  to  9.     What  was  the  stock  of  each  ? 

.  Ans.  A's  |300  and  ^s  $350. 


390  PROBLEMS  IN  PROPOETION. 

16.  What  are  the  two  parts  of  14  of  which  the  greater  divided  by 
the  less  is  to  the  less  divided  by  the  greater  as  16  to  9. 

Ans.  8  and  6. 

16.  In  a  mixture  of  rum  and  brandy,  the  difference  between  the 
quantities  of  each  is  to  the  quantity  of  brandy  as  100  is  to  the  num- 
ber of  gallons  of  rum  ;  and  the  same  difference  is  to  the  quantity  oi 
rum  as  4  is  to  the  number  of  gallons  of  brandy.  How  many  gallons 
are  there  of  each  ?  Ans.  25  gallons  of  rum  and  6  of  brandy. 

lY.  There  is  a  number  consisting  of  three  digits,  the  first  of  which 
is  to  the  second  as  the  second  is  to  the  third ;  the  number  itself  is 
to  the  sum  of  its  digits  as  124  to  7  ;  and  if  594  be  added  to  it 
the  digits  will  be  inverted.     What  is  the  number  ?  Ans.  248. 

18.  A  corn-factor  mixes  wheat  which  cost  10s.  a  bushel  with 
barley  which  cost  4s.  a  bushel,  in  such  proportion  as  to  gain  43f 
per  cent,  by  selling  the  mixiure  at  lis.  a  bushel.  What  is  the  pro- 
portion ?  Ans.  14  bushels  of  wheat  to  9  of  barley. 

19.  What  two  numbers  are  those  whose  sum,  difference,  and  pro- 
duct, are  as  the  numbers  3,  2,  and  5,  respectively.     Ans.  10  and  2. 

20.  What  two  numbers  are  those  whose  sum,  difference,  and  pro- 
duct, ai*e  as  the  numbers  s,  d,  and  p  respectively  ? 

Ans.  — ^and 


s  +  d         s—d 

21.  There  are  two  numbers  in  the  proportion  of  i  to  |,  and  such, 
that  if  they  be  increased  respectively  by  6  and  5,  they  will  be  to  each 
other  as  I  to  i.     What  are  the  numbers  ?  Ans.  30  and  40. 

22.  There  is  a  number,  the  sum  of  whose  digits  is  to  the  humber 
itself  as  4  to  13  ;  and  if  the  digits  be  inverted,  their  difference  will  be 
to  the  number  expressed  as  2  to  31.     What  is  the  number? 

Ans.  39. 

23.  The  difference  of  the  cubes  of  two  numbers  is  to  the  cube  of 
their  difference  as  61  is  to  1,  and  the  product  of  the  numbers  is  320. 
What  are  the  numbers  ?  Ans.  ±20  and  ±16. 

24.  The  sum  of  the  cubes  of  two  numbers  is  to  the  difference  of  their 
cubes  as  559  to  127,  and  the  square  of  the  first  multiplied  by  the 
second  is  294.     What  are  the  numbers  ?  Ans.  7  and  6. 

25.  The  difference  of  the  cubes  of  two  numbers  is  to  their  product 
multiplied  by  their  difference  as  7  is  to  2,  and  the  sum  of  the  numbers 
is  6.     What  are  the  numbers  ?'  Ans.  4  and  2. 


PEOBLEMS   IN   PI^OPORTION.  89i 

26.  The  difference  of  two  numbers  multiplied  by  the  first  is  to  tbe 
difference  multiplied  by  the  second  as  3  to  7,  and  the  first  multiplied 
by  the  square  of  the  second  is  14*7.     What  are  the  numbers? 

Ans.  3  and  7. 
2*7.  The  sum  of  the  squares  of  two  numbers  is  to  the  difference  ot 
their  squares  as  1*7  is  to  8,  and  the  first  number  multiplied  by  the 
square  of  the  second  is  45.     What  are  the  numbers  ? 

Ans.  5  and  3. 

28.  The  sum  of  two  numbers  is  to  the  difference  of  their  squares  as 
1  is  to  4,  and  the  product  of  the  numbers  is  21.  What  are  the  num- 
bers? Ans.   ±7  and   ±3. 

29.  The  sum  of  three  numbers  in  geometrical  progrcvssion  is  62,  and 
the  sum  of  the  extremes  is  to  the  mean  as  10  to  3.  What  are  the 
numbers  ?  Ans.  4,  12,  and  36. 

30.  The  difference  of  two  numbers  is  to  6,  as  the  greater  is  to  the 
less,  and  as  their  sum  is  to  42.     What  are  the  numbers  ? 

Ans.  32  and  24. 


31.  Given   -=b  to  find  x  by  changing  the  equation  into 

dz2aVb 

a  proportion,  Ans.  x=—z . 

^    ^  6  +  1 


««    /-.♦  i/a-i-x-{-  k/a—x     ,«,,,., 

32.  (jiven  ■ — =p  to  find  x  by  changing  the  equation 

into  a  proportion.  Ans.  x=— . 


33.  Given  — r= =2  to  find  x  by  chanepnff  the  equation 

Vx+l-Vx-1  ^  ^    ^  ^ 

into  a  proportion.  Ans.  a;= ||. 

34.  Given    ^  — ^-^==9  to  find  x  by  chanofina:  the  equation 

^4x-{-l-^4x  J  s    B  H  , 

into  a  proportion.  Ans.  x=±. 

35.  Given —b  to  find  x  by  changing  the  equa- 

^      .  ,                 ,.                                 .            a{dol-V2b^b' 
tion  into  a  proportion.  Ans.  x=: . 

^2b-b^ 


CHAPTER    XVIII. 
SERIES. 

(411.)  There  are  many  kind  of  series.  Arithmetical  and  geomet- 
rical series  have  already  been  discussed.  A  series  may  result  from 
dividing  the  numerator  of  a  fraction  by  its  denominator,  or  from  in- 
volution or  evolution. 

•  ^     «<     ♦    >.     »N 


THE    EXPANSION    OF    SERIES. 

(412.)  Quantities  or  algebraic  expressions  may  be  expanded  or 
developed  in  four  ways ;  namely, 

The  Method  of  Division, 
"         "       "   Undetermined  Coefficients, 
"         "        "  Involution, 
and     "         "       "  Evolution. 


THE    METHOD    OF    DIVISION. 

PROBLEM 

1.  Expand into  a  series. 

SOLUTION. 

By  dividing  1  by  1  -I- a,  we  obtain  1— a  4- a'— a' -J-,  <fec. 
14-a)l       (1— a  +  a'— a'  +  ,  &c. 
\-\-a 


—a 

—a- 

-a' 

a' 

a'-^-a* 

-a» 

-a»- 

-a* 

a\ 

THE  METHOD  OP  DIVISION.  898 


Therefore, =1— a +  «'—«'+«*—«'+ a",  &c. 

1  +a 

PROBLEM 
1  1 


2.  Expand  -=- — -  into  a  series. 
^        8     1  +  1 


SOLUTION. 

1 


We  see  that  this  fraction  is  the  same  as if  we  suppose  a  to  be 

1 ;  therefore, 

i=l-7 +  49-343,  A:c. 

8 

The  student  may  not,  at  first  sight,  see  how  this  series  can  be  equal 

to  -.     But  if  the  remainder  is  taken  into  consideration,  the  result 
8 

may  be  verified.    By  considering  the  remainder,  we  have 

ll_7 +49-343+ ?^^ 
8  8 

1         „„„      2401 
__300+  — , 

1_ —2400 +2401 

8  8  '  i 

1=1 

8     8* 

This  fraction  may  also  be  expanded  into  a  finite  series. 
_         1     1.000        ^^^    1     1000      125       125 

^^^'  r  T-=  * ''' '  r  8000=1000=1000=  •  '^'' 

This  method  furnishes  an  explanation  of  the  arithmetical  process 
of  converting  |  into  a  decimal  fraction. 

EXAMPLE  s 

1.  Expand into  a  series. 

Arts.  l+a  +  a'  +  a'+a*+o'  +  a*+ 


«.    -r,         ,  1— a  . 

2»  Expand mto  a  senes. 

^        1+a 

Am.  1  — 2a  +  2a''— 2a'  +  2a*— 2a'  + 


894  THE  METHOD  OF  DIVISION. 

3*  Expand  ^=00  into  a  series. 

Ans.  l  +  l  +  l  +  l  +  l  +  l  +  l  +  l  +  l-l-  .  .  . 

4*  Expand  - — -  into  a  series. 

JL  —  Jj 

Ans.  1+2  +  4  +  8  +  164-32  +  64+  .  , 

5*  Expand 7  into  a  series. 

•^         a—b 

Ans.  i+-  +  _+_  +  _  +  _+   .  . 
a      a       a       a       a 

6«  Expand into  a  series. 

Ans.  a  +  ar  +  ar^  +  ar*  -har*  +  ar^-{-  .  . 


7,  Expand =  into  a  series. 

Ans.  1  —  +-      +     -_  + 
a     a      a       a       a 

1  +x 

8.  Expand into  a  series. 

J.  "~"«C 

Ans.  l  +  2a;  +  2«'  +  2a?*  +  2a;*  + 

9t  Expand  — •  into  a  series. 

l—x  +  x^ 

Ans.  l+a;— a;'— a;*  +  a;'+a;'— ar"— ar'"*  + 
10*  Expand  1  into  a  series. 


Ans. 


x+x  +  x-\-x-\-x+x-{-x  +  x  + 
x—x^  +  x'—x*  +  x''-x''^x''—x'+ 

1-1+1-1+^- J_+i__ 


Note. — ^These  results  may  all  be  verified  if  the  remainders  are  considered 


THE  METHOD  OF  UNDETERMINED  COEFFICIENTS. 

(413.)  The  method  of  undetermined  coefficients  is  a  method  by 
which  algebraic  fractions  may  be  expanded  or  developed  into  a  series, 
the  terms  of  the  series  being  arranged  according  to  the  ascending 
powers  of  one  of  the  quantities  considered  as  a  variable.  This  method 
is  based  upon  the  following  theorems. 


THE  METHOD  OF  UNDETERMINED  COEFFICIENTS.       895 

THEOREM     I. 

(414.)  1.  If  the  series  A3(f  +  Bj^-\-Cx'  +  Dx^-{-,  &c.=^V  + 
B'a^'  +  C'x"'  +  Dx^'  + ,  &c.,  for  all  possible  values  of  x,  the  exponents 
and  coefficients  being  finite  quantities,  and  the  exponents  of  x  in  each 
member  being  arranged  in  the  order  of  their  magnitudes,  commencing 
with  the  least,  then  a=a' ',  6=6';  c=c' ;  d—d'\  &c.,  and  A=A' ; 
B^B'-,  C=C'',  D=D';  &c.    * 


DEMONSTRATION. 

Dividing  both  members  by  xf,  the  equation  becomes  A+Bx^  + 
Cx'-^  +  Dx^  +  ,  &c.=^V-"+^V'-*+<7V-'*  +  i>'a:'"~"  +  ,  &c. 

Since  this  equation  is  true  for  any  value  of  x,  we  have  a  right  to 
suppose  X  equal  to  zero.  Making  this  supposition  the  first  member 
reduces  to  A.  If  we  suppose  all  the  exponents  in  the  second  member 
to  be  finite,  it  would  reduce  to  zero  when  x  equals  zero,  and  we  should 
have  A—0,  which  is  impossible,  as  ^  is  supposed  to  seme  finite  '^^ 
quantity.  Therefore,  it  would  be  improper  to  consider  all  the  ex- 
ponents in  the  second  member  as  finite,  and  hence,  one  or  more  of 
them  must  be  zero.  Since,  a',  b',  c',  d',  &c.,  are  all  different,  it  is  evi- 
dent that  a  can  not  equal  more  than  one  of  these  quantitjps,  and, 
therefore,  of  the  exponents  a'— a;  6'— a;  c'—a-,  d'^a-,  &c.,  but  one 
can  equal  zero. 

It  now  remains  for  us  to  decide  which  exponent  it  is  that  equals 
zero.  Suppose  it  to  be  b'—a,  or  the  exponent  of  x  in  the  second 
term,  and  then  for  x  equal  zero  all  the  terms  after  the  second  must 
vanish,  and  there  would  result  the  equation 

But  since  a:°= 1,  A=A'xf'-^  4-  B. 

Since  a =6'  and  a'  is  less  than  6',  a'—a—&  negative  quantity  which 
put  equal  —n,  and  the  equation  becomes 

AzzzA'x-'^-^rB, 

A=-  +  B, 

A' 
Supposing  x=0  A=-—-\-B, 

A—B^oD. 

But  A—B  must  be  equal  to  a  positive  or  negative  quantity,  or 
zero,  therefore  ^—J5= GO  is  impossible,  which  shows  that  it  is  im- 
proper to  make  b'—a=:0. 

As  the  same  reasoning  applies  to  all  the  exponents  of  x  in  the 


396       THE  METHOD    OF   UNDETERMINED   COEFFICIENTS. 

second  member  but  the  first,  or  a'^a,  it  follows  that  a'— a  must  equal 
zero,  and  the  equation  becomes 

A=A'x\ 
A=A', 
Suppressing  the  equal  terms  A=A'af''~*  the  equation  becomes 
^ar^«+  Cx'-^-\-Dx^  +  ,  &c.,=^V-«+  (7V-«+i>V'-«  +  ,  &c. 
Dividing  by  a:*"*,  we  have 

B+  (7a:'-*  +  i>a;^-*  +  ,  &c.,=i5V'-*+C7V-*  +  i>V-*+,  &c. 
Then  for  a;  =  zero,  we  have  when  we  make  b'—b~  zero,  or  b'=b 

B=B'x\ 
£=B'. 
Thus  we  may  go  on  and  prove  that  c=c'  and  C=zC' ;  d=d'  and 
J)==:D\  (fee,  which  proves  the  theorem. 

THEOREM    II. 

(41 5.)  If  Aaf-\-Bx^+  Cx'  +  Dxf^-h,  <fec.  =0  for  all  possible  values 
of  X,  a  being  less  than  6,  b  less  than  c,  c  less  than  c?,  &c.,  each  of  the 
coefficients  must  be  equal  to  zero. 

DEMONSTRATION. 

Dividing  by  ar",  we  have 

"      A  +  Ba^"  +  Cx'-^  +  T>:x^-^  + ,  &c.  =0. 
Know  we  make  x  =  zero,  we  obtain 

^   ^  =  0. 
Suppressing  ^=0,  and  dividing  by  a^,  we  obtain 
B  4-  Cx'  -*  +  Dx^^  + ,  &c.  =0, 
in  which  if  a;  is  put  equal  to  jsero  the  equation  reduces  to 

In  the  same  way,  we  can  get  (7=0,  i)=0,  <fec.,  which  was  what 
was  to  be  proved. 

Let  us  now  apply  the  method  of  undetermined  coefficients  to  the 
expansion  of  series. 

PROBLEM 

1 


(416.)    1.  Expand  j-^^-^,  into 


a  senes. 


SOLUTION. 


By  an  inspection  of  this  fraction  we  see  that  by  division  the  first 
term  of  the  quotient  would  not  contain  x  but  the  second  term  would. 


THE  METHOD   OF   UNDETERMIKED   COEFFICIENTS. 
1 


397 


Assume,  chen, 


^A  +  Bx+Cx'  +  Dx^  +  Ex'-^. 


l  —  ^x+x" 
Clearing  of  fractions,  we  have 

1=^+  Bx+    Cx^+  Dx'+  Ex'+,,.,. 

--2Ax-2Bx^-2Cx'-2Dx'- 

4-   Ax'+  Bx'+    Cx'-\- 

Transposing,  the  equation  becomes 

\+2Ax  +  2Bx''  +  2Cx^  +  2Dx'-\- =A  +  Bx+{A+C)3^-\- 

{B  +  Dy-{-{C+E)x'-\- 

According  to  Theorem  1  (414),  the  corresponding  coefficients  in 
these  series  must  be  equal. 

A=\    ]  r  ^=1 

B=2A  B=2 

A+Cz=2B  ,  C=S 

B  +  I)=2C    P^^^^^   1    D=4 


C+E=2D 
&c. 


&c. 


Substituting  these  values  in  the  assumed  series,  we  have 


*     \-2x  +  x' 

This  fraction  might  have  been  developed  into  a  series  with  more 
fecility  by  division. 

The  student  should  be  very  careful  in  assuming  the  series  of  unde- 
termined coefficients.    This  may  be  illustrated  by  the  following 


PROBLEM 


2.  Expand 


2 


3a;'— 2a;'* 


-r  mto  a  senes. 


SOLUTION 


By  division  we  see  that  the  first  term  of  the  expansion  must  be 


2 


2       2 
= — 5=-a;~',  therefore,  assume 

OX         O 

■=Ax-^  +  Bx-'  +  Cx'  -\-Dx+ Ex*  -f  Fx^ + &c. 


3a;'— 2a;' 

Clearing  of  fractions  and  transposing,  we  have 
2  +  2^a;  +  2J?a;»+2Ca;'  +  2i)a;*4-2^a;»  +  ,  &c.  =3-4+3^a;  +  3(7a;'-|- 
3i>a;' +  3^ar*  +  3i^a;' +  <fec. 


398        THE  METHOD   OF   UNDETERMINED   COEFFICIENTS. 


By  equating,  like  coefficients, 


we  have 


C   8A=2       ] 

r  -4=f 

3B=2A 

^=f 

S0=2B 

c=^s 

3D=2C    y,  whence   - 

I>=H 

3E=2D 

^=2% 

3F=i2E 

F=^%\ 

&c. 

&c. 

By  substituting  these  values  in  the  assumed  series,  we  get 
2 


4       8      16a;     32.tr'     64a;''       , 


8a;'— 2a;«     Sx^     9a;     27      81      243      729 

2.2         2'      2'     2*a;  ,  2V     2V       »         ,  .i.    i         i? 

3^^=2?=3-?+3V+§^+ 3^+T^-»-T«-+  ^^-^  ^^^"^^  '^^  ^'^  '^ 
the  terms  is  apparent. 

2 


Since 


assuming  it  equal  to  ^  +  ^a;  +  Cx^  +  Dx^  +  &c.,  and  then  multiply- 

2  2 

ing  the  result  by  — ^  for  the  development  of  — -^ — --|. 
a;  oaj  ~^JiX 

o 
PROBLEM 

3.  Expand  (1  +a;)i  into  a  series. 

SOLUTION. 

Assume  (H-a;)i=^  +  5a;+(7a;'  +  i>a;'  +  ^a;*+ 

Squaring 

l+a;+0a;'-K)a;''+0a;*+&c.  =-4'+2^J?a;+2^Ca;'  +  2^i>a;'  +  2^^a;*+ 

^V  +  2BCx'  +  2^i>a;*+ 

+  (7V+..:.. 

Equating  like  coefficients 


^'=1    ^ 

2^J?=1 

2AC+B''=0 

2AD  +  2BC=0 

2AE-^2BD-hC^=0 

&c. 


>•,  whence   - 


A=l 
B=i 


we  have   " 

Substituting  these  values  in  the  assumed  series,  we  get 


0=-i 
&c. 


♦/l+a;=14-ia;-ia-'  +  yV^«-yf^a;*  + 


DECOMPOSITION  OF  RATIONAL  FRACTIONS.  899 

But  we  had  a  right  to  assume 

Vl+x=-A-Bx~Cx''—I)x'—JEx*— 

By  squaring  we  should  obtain  the  same  equation  as  before,  and 
hence  the  same  values  for  A,  B,  (7,  i>,  E,  <fec.,  which  being  substituted 
in  the  assumed  expression  would  give 

VT:^x=-1-^x  +  \x''-^\x'+j^^x'-  ...... 

From  these  two  results  we  see  that 

VT+x=±.\±^x^z\x'^±^\x'^jijx'± 

the  upper  row  of  signs  being  taken  for  one  result  and  the  lower  row 
for  the  other ;  or  we  may  write  it  thus, 

V\+i=  ±  (1  +\x-\x'  +  -^^x'-^^^x'  + ) 


DECOMPOSITION    OF    RATIONAL    FRACTIONS. 

(417.)  The  method  of  undetermined  coefficients  may  be  em- 
ployed in  the  decomposition  of  rational  fractions,  and  for  this  purpose 
it  is  often  used  in  the  Integral  Calculus. 

PROBLEM. 

5a;— 19 


Decompose  the  fraction 


x'—Sx  +  lb 

so  LUTION 


_  6a;— 19  5a;— 19        ,  ^  ^,   ^ 

Since  -= — ^  ■,  ^^  —7 -r-. -.  let  us  assume  that 

a;"— 8a;+15      (a;~3)(a;— 5) 

5a;— 19       _  A  B 

(a;— 3)(a;— '5)~a;— 3     a;— 5 

5a;— 19      _A{x—b)+B{x—Z) 

(a;-3)(a;.— 5)~       (a;— 3)(a;— 5) 

5ar— 19=^(a;— 5)  +  ^(a;— 3) 

(6^  +  3^)a;"  +  6a;=  iga;"  -\-{A-{-  B)x 

Equating  like  coefficients  of  a;, 

We  have    5^^+3^=19, 
(    A-{-  B=5, 
5^  +  3^=19, 
3^  +  35=15. 


2Az 

=4, 

A-. 

=2, 

Bz 

=  3. 

6a;— 19 
a;"  _- 8a- -1-16 

X 

^^- 

3 
—6' 

400  DECOMPOSITION   OF   RATIONAL  FRACTIONS. 

Note. — The  values  of  A  and  B  might  have  been  determined  in  a 
different  manner.     The  equation 

6x—l9=A(x—5)+B{x—S). 

Since  this  equation  is  true  for  all  values  of  ar,  let  us  assume  it  to  be 
3,  and  we  have         15  — 19  =  — 2^,  or  ^  =  2. 

If  x=5  we  have  25  —  19=     2B,  or  ^=3. 

EXAMPLES. 

1.  Expand mto  a  senes. 

L  -{''^  r  X 

^ws.  1— 2a;  +  a;»4-a?»— 2ar*-t-a;'+a?*— 2a;'+ 


2*  Expand  — r-  into  a  series. 
^        c  +  bx 

.        a     ah       ab^    -     ab^  .     ab*  . 
Am.  -——^x-\ — r-ar' ^^  -^  -t» — 

3t  Expand  -, rr  into  a  series. 

^        (a—xy 

1    .  2a:  .  3a;'     4a;'  ,  5a;* 

Am.  —4-  —+—  +—  +  --4-  .  . 

a"     a^       a*       a^       a' 

.    ^         ,  a—bx  . 

I*  Expand mto  a  senes. 

^        a  +  cx 


Am.  l-(b  +  c)-  +  c(b^c)~-c%b-\-c)^,+ 

(t  CL  0/ 


o«  Expand  - — —  mto  a  series. 

^         1  — 2a;— 3a;' 

Am.  l+a;  +  5a;'  +  13a;'  +  41a;*  +  121a;'  +  365a;"+  .  .  . 

6i  Expand into  a  series. 

b—ax 

d  ,  adx  ,  a^dx^  ,  c^da^  ,  a^dx* 
Ans.  -  +  _-  +  _-+-^+-_+ 

7«  Expand  (a—x)~^  into  a  series. 

Am.  a-*+a~'a;4-a~V  +  a~V4-a~V  + 

a^—x^ 

8.  Expand into  a  series. 

a—x 

Am.  a*+a*x-{-a^si!'-^ax*-\-xl^. 

bx 
9i  Expand  — -^— r  into  a  series. 
^        ax-\-bx^ 

Am.  -——rX-\ — ra;' ^x^-\- 

a      a"        or         a* 


10*  Expand into  a  series. 

a;-|- 1 

Am.  a;— a;' +  a;'— a;* +«'—«;•  4-, 


DECOMPOSITION  OP  RATIONAL  FRACTIONS.  401 

lit  Expand into  a  series. 

^        a:--l 

Ans.  —  1  — 2a;— 2a;'~2a;'— 2a;*— 

12i  Expand  •- into  a  series. 

^         1— a;— ic' 

Ans.  H-2a;+3a;'  +  5a;'  +  8a;*  +  13a;*  + 

13.  Expand  — into  a  series. 

^         1+ic  +  a;' 

Ans,  x—x'^-\-x*—x^+x''— 

14,  Expand  a;(l— a;)"^  into  a  series. 

Ans.  a;+3ic'  +  6a;'  +  10a;*+  .... 

15*  Expand  ^a^—x^  into  a  series. 

x'       X*        x'         5x^ 

Ans.  a — - — ;— ' r— 

2a      8a'      16a'      128a' 

16(  Expand  4/64  + 1  into  a  series. 

Ans.  8  4-rr;:— -ri  +  r-r-. 

16      8*     2-8* 

17,  Expand  (a+x+x^)-k  into  a  series. 

Ans.  ai+^  +  /--— ^W 

2a2     \2a2     Sa^^ 

18.  Expand  (a^—ax+x^)-h^  into  a  series. 

X      Sx" 


Ans.  a— -4-- 

2      8a 


19.  Expand  (ax—x*)i  into  a  series. 


J.  1  •    a;  2    ,    a;^ 


Ans.  a^x^— — j-\-       , 
2a2     8a2 

20.  Expand  a(a'  +  6')-i  into  a  series. 

,       ,       b'    ,   26"      146» 
^^^-  '-^  +  -9^-^!^ 

21.  Expand -^  into  a  series. 

Ans.  zr-  +  -  +  77z:X  +  -—x'-^    .   .   .  . 
3a;     9     27       81 

„    _  10a;-16  .  3,7 

22.  Decompose  j,-^^^.  ^      ^«,.  __+__. 

•»   Tx  lla;-37  ^5^6 

23.  Decompose  ^._^^^,q  Ans.  j^j:^ 

24.  Decompose  -5 — — — ;.  Ans.  -— -. 

^       a;'— 12a;  +  32  a;— 4     a;— 8 

20 


402  THE  METHOD   OF  INVOLUTION. 

25.  Decompose  ——---—---,        Ans.  _—-  +  ---  + 


x^  —  ex'+llx—G  x  —  1     x—2     x—3 

26.  Decompose  — j — -  Ans.  — r\~77 — rr\~o/  a  ■  i\- 

«.•    -r^  ar+2  ,  1,32 

27.  Decompose  — 5 Ans.-—, — — :x+-z7 7t~- 

28.  Decompose 


^715.   -—, --^  — + 


2(fl;+l)      x  +  2     2(a;i-3) 

^^    X.  2a;  +  3  ,  3  1,5 

29.  Decompose     3  ,    , — --.      Ans.  — — —  .,,    ,  ^^+-r? ^• 

^        x^+x^—2x  2x      6(x  +  2)      3(ar— 1) 

QA    T.  13  + 21a; +  23?" 

30.  Decompose    ^_g^.^^^.  ■ 


Ans. — +- — r-  +  - 


1+x     1—x     l+2a;     1— 2a; 


THE    METHOD    OF    INVOLUTION. 

PROBLEM. 

(418,)  Expand  (a;+y)"  into  a  series,  n  being  a  whole  nmnber. 

SOLUTION. 

This  may  be  expanded  by  raising  a;  +  y  to  the  nth  power. 

This  may.  be  done  by  actual  multiplication,  or  by  the  binomial 
theorem.  As  this  theorem  is  also  applicable  to  evolution,  it  will  be 
demonstrated  in  the  next  method. 


THE    METHOD    OF    EVOLUTION. 

(419t)  A  method  of  extracting  roots  has  already  been  given. 
But  since  th«  law  of  the  binomial  theorem  holds  good  when  the  ex- 
ponent is  fractional,  we  have  a  more  expeditious  method  of  extracting 
the  roots  of  binomials. 

BINOMIAL     THEOREM. 

4?*— 1)  .  o    1     n(n—l)(n—2)  _,  . 
(x  +  i/Y  =  a;"  +  nx"- '  y  +  -^ — ^a;"-^  y'  +  "g  3      ^      ^  "^ 

■A  ""   ^^ 1^ -x'^*y*-\-   &c.     X  and  y  being  any  quantities, 


THE  METHOD  OF  EVOLUTION.  403 

whether  real  or  imaginary,  and  n  being  a  positive  or  negative  integer, 
or  a  positive  or  negative  fraction. 

DEMONSTRATION. 
This  demonstration  depends  upon  the  following 

LEMMA. 

rws"*"^,  n  being  a  positive  integer  or  a  positive  fraction, 


\  z—u  A 


a  negative  integer  or  a  negative  fraction.  The  notation  after  the 
parenthesis  denotes  that  the  first  member  of  the  equation  equals  the 
second  when  u  is  equal  to  z. 

We  may  divide  the  proof  of  this  lemma  into  four  cases. 

CASE   I. 
When  w  is  a  positive  integer. 

PROOF. 

We  have  seen  in  (112)  that  when  w  is  a  positive  integer  that 

z—u 
When  u=zz  each  of  the  terms  in  the  second  member  becomes  g""*, 

I  = 

wg''"^  which  proves  the  lemma  true  in  the  first  case. 

CAS.E    II. 
When  72  is  a  positive  fraction. 

PROOF. 

Supposmg  n=r-^  p  and  q  being  positive  integers,  and  a=r',  and 

u=zv'^^  we  have 

£.        E  

Zq  —  uq     rP—vf      r—v 


.u       r'^  —  v'^     r''—v'' 


fP  —  '\)V  rf.q — 2/7 

According  to  Case  I. —pr^-^,  and =qr^^.  when  t>=r. 

°  r—v  r—v 

But  since,  when  vz=r^u  must  equal  s,  we  have 
L       t 

(Zq—Uq\       _ 
Z—U   /^~ 


'rP- 

-?n 

r- 

-V 

3IMF1 

nam 

7-7  _ 

-V^ 

r— 

-V 

^- =^-rP~^^^ 

qr^-'     q 


—I. 


404  THE  METHOD  OF  EVOLUTION. 

Butsiiicer'=2,  wehaver=2j  andr^^=2~y  =2*    '*     Substituting 

— ^1      = 

z—u  /«^, 

^  y       which  is  the  formula  ffiven  in  the  lemma,  when  »=-,  and  hence 

proves  the  lemma  to  be  true  in  the  second  case. 

CASE     III. 
When  w  is  a  negative  integer. 

PROOF. 

Suppose  n=—m.    Then  since  z~"*—u~^=^g~^u~^(z'*—u^)j  we 

liave =  —zr^u-^{ 1. 

z—u  \  z—u  / 

j      r=mj^\  which 

1      = 

_^-»^-«i,yjg«-i___^2-m-i^  which  is  the  formula  given  in  the  lemma 
when  «= — wi,  and,  therefore,  proves  the  lenuna  to  be  true  in  the 
third  case. 

CASE    IV. 
When  w  is  a  negative  fraction. 

PROOF. 

P 

The  proof  in  this  case  is  found  by  putting  —  for  —m  in  the  last, 

and  referring  to  Case  11.  instead  of  Case  I. 

Therefore,  the  truth  of  the  lemma  is  fully  established. 

We  are  now  ready  to  proceed  with  the  general  demonstration  of 
the  binomial  theorem.  We  are  required  to  ascertain  the  development 
of(a;4-y)^ 

Since,  (aJ+y)"=a^"(l+-)  it  is  only  necessary  in  order  to  find  the 

development  of  {x-\-yY  to  find  that  of  (1  +-)   and  multiply  it  by  a?*. 

For  convenience  put  2=-,  and  we  then  have(l  +zY,  whose  develop- 
ment will  now  be  sought. 


THE  METHOD  OF  EVOLUTIOlf.  405 

Assume  (1-\-zY=A-^JBz-{-Cz^  +  I)z^  +  jE:z*+ in  which 

the  assumed  coefficients  A,  B,  C,  B,  JS,  &c.,  are  independent  of  2,  and 
depend  entirely  for  their  values  on  1  and  n.  Let  us  now  endeavor  to 
ascertain  the  values  of  these  coefficients  in  terms  of  1  and  n. 

To  find  the  value  of  ^  make  2=0  in  the  assumed  equation,  which 
we  have  a  right  to  do,  since  A  is  independent  of  z,  and  we  get 

Substituting  this  value  of  A  in  the  assumed  equation,  we  have 

{l+zy=l+£z+Cz'-{-Dz'-{-M*-^ {A). 

Since  z  may  be  any  value  whatever  without  affecting  the  coefficients 
Aj  B,  C,  &c.,  let  us  make  z=u^  and  we  have 

(l+uy=l+Bu-hCu''-{-Du'  +  I!u*-{- (B) 

Subtracting  (B)  from  (A)^  we  get 
{l+zy-{l+uY=B{z-u)+C(z'-u')  +  D(z'^u')  +  i;(z*-u*)-{.  .... 
Dividing  the  first  member  of  this  equation  by  (1  +2)— (1  +u),  and 
the  second  by  its  equal  z—u,  we  obtain 

(i-\-z)-(i+u)      ^  \z-ur    \z-ur    \z-ur    ^  ' 

Putting  l+s=r  and  l+w=v  (C7)  becomes 

rlzr^B^  c(^+i>(^^)+i.f-«:)H. (2,) 

r—v  \z—uj        \z^u1        \z—uf 

But  by  the  lemma  we  have 


nr^ 


Lemma 


But  when  ?;=r,  we  also  have  w=2,  and  by  the 

iZ^  —  U^  \ 

\  Z  —  U  /«^, 

Equation  (D)  will  become  by  substitution 

wr*^* = ^  +  2  Cfe  +  3i>2' +  4^*. 
Multiplying  this  equation  by  r= 1+2,  we  get 

wr'»=^  +  2(72  +  3i)2''  +  4^2'  +  . 

-vBz-V2Cz^^-ZDz^-\- 

Since   14-2=r,   we   have   7i(l +2)''=^  +  (^  +  2(7)2+(2C7+3i)) 

2''  +  (3i>  +  4J')2'-f.    .    .    . 

Multiplying  (A)  by  n^  we  get 

w(l+2)"=ri  +  w^0  +  »6fe=  +  /ii)«"  +  .  .  .  . 


406 


THE  METHOD  OF  EVOLUTlOlT* 


Whence,  ^  +  (^  + 2  (7)2 +(2(7+ 3i))2».f(3i>  + 4^)2'  +  .  .  .  . 

=n-hnBz  +  nCz''  +  nBz^  + 

Equating  the  coefficients  of  the  like  powers, 

f  B=n, 
B=n       > 


we  have  < 


B  +  2C=nB, 

3i>  +  4^=wi>, 

(fee. 


► ,  whence    < 


C= 


D= 


n[n—l 


n(n—l){n  —  2) 


3 


n(n-l){n-2)(n-S) 
2.3.4      ' 
&c. 

Substituting  these  values  of  B,  (7,  2),  &c.,  in  (A)^  and  restoring  the 
value  of  2=-,  we  get 

w(»— 1)(%— 2)(w— 3)     y* 
■^        2.3.4         ^"^^  •  • 

Therefore,  since  a;"(  1 +-J  =:(ar  +  y)'',  we  get  by  multiplying  the  ex- 
pression of  1 1  +  - 1  by  a;", 

^(^—1)      o  o     w(w— l)(w— 2)      ,  . 
.       (ar+y)"^.g"  +  naf-V+    ^  ^     ^^"^y+        2  3  "^ 

n{n-\){n-2){n-3) 

^       2.3.4''     ^"^ 

which  is  the  formula  given  in  the  theorem. 

For  the  purpose  of  reference,  we  append  the  following  formulas 
which  have  been  encountered  in  the  above  investigation.  Which 
formula  it  is  best  to  use  will  depend  on  the  nature  of  the  binomial  to 
be  expanded. 


(a; + y)"=ar'*  +  Tiaf^^y + ^^^^— -Wy 


,(n-\){n-2)^_, 


af-y  +  .  .  (1). 


THE  METHOD  OF  EVOLUTION.  -      *   407 


EXAMPLES 

1.  Expand  (a^+x)-k  into  a  series. 


X         x^      ,      3a;'  3-5a;* 

Ans,  a-\-—-————L'r 


2a      2-40"      2-4-6a'     2-4-6-8a* 
2.  Expand  (a+a:)i  into  a  series. 

X         x"     .      3a:'  S'bx" 


.  ain 


^715.  ai(H- 


2a     2'4a'     2-4-6a'      2-4-6-8a* 
3«  Expand   4/2  =  (1  +  l)i  into  a  series. 

^^^-  l  +  2"2^4  +  2^"T4^+   • 
^     4»  Expand  (a— 6)i  into  a  series. 
p  .         J^      b        35''        3-76'  3-'7-ll6* 


bi«.  ail 


r 


4a     4-8a'     4-8-12a'      4-8-12-16a*  / 

5«  Expand  (a  +  y)"~*  into  a  series. 

1      4y      10y»      20/     36y* 
^''*-  "^     "^^"^^ a"'""^  "^ 

6*  Expand  7  into  a  series. 

c ^  be ^b\     b'c  ,b*c 
a     (T      a'      a      a° 

7.  Expand  (a'+^'H  into  a  series. 

6»      6*       5- 

Ans.  a  +  — -—--,  +  — —-7 

2a     8a'      16a' 

8.  Expand  —     •       into  a  senes. 

n^a^^x^ 

a/        x^    ,    3a;*    ,     bx"    ,  5.7a;"  \ 

^^-  6('-2^^-2V  ■^-2V  +  2V    •  •  •  •; 

9.  Expand  (c'— a;'')f  into  a  series. 

4/,       3a;'      3a;*      6a;«  \ 

^^^-  n^-2V~2V-2V ; 

lOt  Find  the  cube  root  of  — — ^3. 
a  -f"  0 


"  ^      3a' 

'      9a" 

81a» 

11. 

Expand   4/5  = 

=  1/4  +  1  into 
Ans. 

a  series. 

'4- 

-1  +  i- 

2"      2» 

5 

2" 

12. 

Expand  V6= 

=V8— 2  into 

a  series. 

-4ri^. 

-ji- 

2^3'"^" 

6 

5 

2'-3* 

2"-3*  •  • 

iOQ  THE  METHOD  OV  EVOLtTTIOIir. 

PROPOSITION 

(420»)  1.  The  sum  of  the  coefficients  of  the  odd  terms  of  the  eX' 
pansion  of  {a-\-  hy  is  equal  to  the  sum  of  the  coefficients  of  the  even 
terms* 

PROPOSITION 

(421.)  2.  The  sum  of  the  coefficients  in  the  expansimi  of  {a  +  hy 
is  equal  to  nth  power  of  2. 

PROP  OSITION 

(422i)  3.  The  sum  of  the  coefficients  in  the  expansion  of  {a—hy 
is  equal  to  zero. 

(423.)  We  have  seen  that  the  binomial  theorem  furnishes  the 
means  of  expanding  polynomials,  since  all  polynomials  may  become 
binomials  by  substitution.  But  a  more  direct  method  of  expanding 
polynomials  is  by  the 

MULTINOMIAL  THEOREM. 

{x+ay  +  hy^^-cy^  .  .  )''=t-^nx'^^ay  +  \^-^ 

DEMONSTRATION. 

r 

Assume  {x-\-ay-\-hy^^cy^  ....  )Tz=A-{- By  •{-  Cy^ -\- Dy^ ■{■  . .  (-4). 
To  determine  A  make  y=0.     Whence 

r 

A=x~*^ 

Making  z—y^YfQ  gei 

{x-\-az  +  bz'-\-cz^ )l=:A-\-JBz-{-Cz^  +  Dz*-}- {B), 

Subtracting  (B)  from  (A)  there  results 

{z-\-ay  +  by^-\-cy* )'^'-{x  +  az  +  bz^  +  cz''  .  .  .  )T=^(y— g) 

+  C{f-z')-hD{y'-z')+  ...((7). 
Putting      P* = (.r  +  ay  4-  by'  +  cy' ), 

r 

whence    P''=(x-^ay-{-by'-\-cy^  .  .  .  .)7, 
and  Q*={x^az+bz'-\-cz* ) 

r 

whence       Q'=(x-^az-\'bz'-{-cz^ » •  •  ')~** 

wegetR-Q'=B{y-z)+C{f-z')  +  D{y'-z')-h 

But       P'-^Q'=a{y-z)-hb(f-z')-\-c{y'^z')-^  .... 


Tfifi  MEtHOD   OF  EVOLUTION. 


409 


"Whence,  by  division,  we  obtain  I 

P^-^Q'^B{y-z)+C(f-z^)  +  D{f-z')+  .... 
P'-Q'        a{y-z)-\-b{f^z^)+c{f-z')+ 

Dividing  the  numerator  and  denominator  of  the  first  member  by 
P—Q  and  the  numerator  and  denominator  of  the  second  member  by 
y—z,we  get 


B+C{y  +  z)  +  I){y'  +  yz-\-z')-^  . 


a-{-b(y+z)+C(y''  +  yz  +  z')  + 


W 


Now,  if  we  make  y=z,  since  P  will  then  equal  §,  Eq.  (jD)  becomes 
j.pr-1     y.pr     B-h2Cy-{-SDy''  +  4£Jy'-\-  .... 


sF'-'     sF*       a  +  26y  +  Sc?/"  +  4dy^  +  .  .  .  . 
Substituting  the  values  of  P'  and  P',  we  get 

r(x  +  ay  +  by''  +  cy^  .  .  .  .  )T_^  +  2(7y  +  3i)y'+4^/+  .  .  .  . 
s(x  +  ay  +  by'-{'cy^  .  .  .  .  )   ~  a+2by-{-  dcy""  +  4rfy'  +  .  .  .  . 

Clearing  of  fractions,  we  have 
-(x+ay  +  by^+cy^  +  dy^  .  .  .  y(a-^2by-\-Scy^-\-4dy'  +  5ey*  .  .  .  )  = 
(a;  +  ay4.6y»+cy'-f  rf/ .  .  .  ^B  +  ^Cy  +  SDy'  +  iEy'-hSFy'  .  .  .) 

By  substitution  from  Eq.  (-4),  we  get 

-(A+JBy-{-Cy''  +  Dy'-i-Ey\..){a-h2by  +  3cy''  +  4:dy'-\-5ey\,,)=: 
s 

^x  +  ay  +  by'  +  cy'  +  dy*  .  .  .  ){B  +  2Cy  +  Sl)y'-{-4:Ey'-{-5Fy*  ,  .  .) 
Performing  the  multiplication  indicated,  we  have 


-aA  +  aB 

8 

-y-uiC 

y^aD 

y^B 

y.aF 

-y'+aG 
s 

"^f-vaH 

+  2bA 

+2bB 

+2bC 

-^2bD 

-\-2bF 

+2bF 

+2bG 

+3cA 

+3cB 

+3c(7 

+ScD 

+ScB 

+3cF 

ndA\ 

^4:dB 

+4dC 

-\-4dJ) 

+4:dF 

+5eA 

-\-5eB 

^heC 

■¥5  el) 

+6/4 

WO 
+ShA 

[Forward 


410 


THE  METHOD   OF  EVOLUTION. 


BxMiB  y-\-h  B 

f+cB 

f+dB 

y*+eB 

f^fB 

f+gB 

+2x0 

+2aa 

+2bC 

+2cO 

-v2dC 

+2eC 

+2/0 

^3xD 

+3al) 

■V3hD 

^ZcD 

+SdI) 

+3eD 

+4xB 

+4.aE 

•^^hE 

+4cE 

+4dE 

+5xF 

+baF 

+5bF 

+5cF 

+6xG 

+6aO 

+6bG 

-¥lxH 

+naH 

+SxI 

Equating  the  coeflficients  of  the  like  powers  of  y,  we  have 

T 

Bx^-aA 
s 

aB-{-2xC=aB--{-2hJ!^ 

8                 S 

bB-{-2aO+Sx£>=aC 

-  +  2bB-+3cA- 

S                S                8 

cB-^2bC-^  3aD  +  AxE=aD-  +  2bC~  +  3cB-  +  4dA- 

8  S  S  S 

dB  +  2cC-h3bD  +  4aE+5xF=aE-  +  2bD^  +  3cC--{-4dB-+5eA^ 

s  s  s  s  s 

eB+2dC+3cI)+4tbE+5aF+6xG=aF^2bE-+3cJ)-+idC'^5eB-+efA- 

S  S  S  S  S  8 

There  is  a  symmetry  about  these  equations  which  would  enable  us 
to  fonn  successive  ones  without  resorting  to  the  equation  from  which 
these  have  been  derived. 

r  .  .  L 

Putting  71=-,  and  instead  of  A  write  its  value  x>=afy  and  solving 

the  above  equations,  we  have 

A=af 

JB=nax'^^ 

nin-l){n---2) 


3 


'a^x'^^-\-n{n—l)ahar-^-{-ncaf^^'' 


a^bx^'+'!^\2cu;+b')xr-2 


„^n{n-l){nr-2){nr-3)  n{n~l){n-2) 

2    .    3    »    4  2 

+ndar-\ 

It  may  be  seen  from  these  values  we  are  not  able  to  write  all  the 
terms  in  the  value  of  F,  although  some  are  apparent.  But,  by  using 
A,  jB,  (7,  i),  &c.,  instead  of  their  values,  the  law  becomes  evident. 

Thus,  we  may  obtain  the  following  general  formula  : 


THE  METHOD  OF  EVOLUTION. 


411. 


+  + 

f  I 

«  00 

I  ? 


+ 

I 

OS 

+ 

? 

I 

en 

I 

f 


i: 


? 
? 

? 

I 


f 

CO 


f 

+ 


col- 
I 

+ 


05 

'oo 

I 

Or 

? 


^    +     b 
^     f 

CO  ^ 


to 


to 
+ 

00 


+     + 


to 


3      -i^ 


^ 


CO 

I 

CO 

b 


^       to 

b    Q 

+       4- 


5S 


b 

os 

,    I: 

b    *«s, 

i 
1 


Si 
I 

CO 

b 
+ 

'co 

I 

to 


b 

en 

i 


CO 

+ 

'fco" 

I 

bO 

+ 

'co 


b 

+ 

i 


+ 

wi- 


ts 

to 


b 

09 

i 

^1 


+ 

to 


+ 

+ 

+ 

+ 

+ 


412 


EKVERSION  OF  SERIES. 


MeIemabk. — This  theorem  applies,  of  course,  to  both  involution  and  evolution. 
The  student  should  use  the  multinomial  theorem  in  solving  the  following 

EXAMPLES. 

1.  Expand  {l-h^  +  x'  +  x^  +  x*  .  .  .y  into  a  series. 

Ans.  1  +  3^  +  6a;' 4- 10a;' 4-1 5a;*  + 

2«  Expand  (2a;  +  3a;' + 4a;'*  ....)'  into  a  series. 

Ans.  8a;'  +  36a;*  +  102a;'^  +  231a;'+ 

3t  Expand  (1-1- a;  4- a;' +  a;' +  a;*  .  .  .  .)i  into  a  series. 

Ans.  l  +  Xa;-}-fa;»  +  y\a;'  +  -i?jLa^*-h 

4*  Eicpand  (l-f-^a;  +  ^a;''+ia;'  .  .  .)i  into  a  series. 

Ans.  l+aa;  +  J^^'-f-eV8«'  +  «\¥7^*+ 


REVERSION  OF   SERIES. 


(424.)  The  reversion  of  a  series  is  finding  another  series  which  is 
equal  to  the  unknown  quantity  in  the  given  series. 

PROBLEM 

(425.)  1.  Revert  the  series  aa; -I- 6a;' 4- ca;'H-6?a;* 

SOLUTION. 

Assume  that  x=Ai/-{-By^+  Ci/^  +  Dy* ,  y  being  equal  to 

ax-\-bx^-hcx^-{-dx*'  .... 
Substituting  the  value  x  in  the  equation 

yzzzax  +  bx'  +  cx^  +  dx*  .  .  .  ., 
we  have 


y^Of  +  0i/*  +  Oy*=aAy+aB  -' 

hA 


■\-2hAB 
cA' 


Equating  the.  like  coeflScients  of  y, 


we  have  aA^V 

6^'4a^=0 

cA^  +  1hAB+aG=0 

dA*+ScA'B+hB'+2bAC+aD=:0 


.    whence 


y'4-        aD 

4-  2b AG 

+        bJB' 

-\-3cA'JB 

dA 


^4 

a 

or 
^^26'-ac, 


y*+  .... 


I 


B=- 


bh'—babc+a^d 


1        b    ,      26'— ac   3     bb''—5abc-\-a^d  , 


Therefore,  a;=-y jy'  -I g — y 

CL        a  a 


REVERSION  OF  SERIES. 


m 


PROBLEM 

2.  'ReveTti/=ax+hx^+cx^+dz''  .  .  .  , 

SOLUTION. 

Assume  x—Ay  +  By^  +  Cy^ + JOy''  .  . 
Substituting  x  for  its  value,  we  have 


y+Oy'  +  Oy'  +  0''=aAy-\-aB  f-\- 

aO 

y'+       aD 

/+..•. 

hA' 

+  dhA'B 

+  3M'(7 

cA^ 

■\-hcA^B 

dA' 

+  3bAJS' 

Equating  like  coefficients,  we  obtain 

\a==\ 

- 

a' 

aA=l  ' 

Cb 

■' 

bA'  +  ajB=0 

4) 

cA'  +  ShA'J3  +  aC=0 
SlAB'+dA'+5cA'B+BM^C+aD=0  J 

■ ,  whence   • 

a 

-ac 

7          > 

jy__  I2h-8dbc+cfid 

a" 


PROBLEM 

3.  "Revert  ax +hx'*+cx^-\-dx*  ....  =py-\-qy*-{-ry*+sy* 

SOLUTIO  N. 

Assume  y= Ax -{■  Bx^  +  Cx^-\-Dx*  .... 
Substituting  this  value  of  y,  we  have 


ax  +  bx^+cx^+dx* =pAx+pB  |a;'4-       pC 

qA""]     +2qAB 
rA' 


Equating  the  like  coefficients  of  ar,  there  results 

a 


a;»+       pl> 

+  2qAC 

+        qB' 

-hSrA'B 

sA' 


pA=a' 

qA^+pB=b 

rA^+2qAB+pC=c 

8A*+3rA^B+q&+2qA  C+pD=d 


,  whence  ■ 


P' 


j^_bp^-a*q 

ri_cp*—2abp'^q  +  a^q + 2cfipr 

J'  • 

rj_cp*—a*pf  +  2aq(a^q—hpV 


Therefore,  y^g^  +  ^X:zg!g,.+'^'-">  +  y  g:±l)^.+  .  .. 
P  P  P 


414  REVERSION  OF  SERIES. 

PROBLEM 

4.  Revert  ^+aa;  4- 6a? + car* +  c?a;* .  . .  =y. 

SOL0TIO  N. 

Transposing  jt?,  we  have 

ax  +  bx-\-cx^-\-dx*  .  .  .  z=y—p, 
=  This  series  is  the  same  as  the  one  in  the  first  problem  except  that 
y  — p  stands  in  the  place  of  y.    Therefore,  its  reversion  must  be. 

5b'—5abc-\-a'd, 

— ^ — ^-^y 

EXAMPLES. 

L  Reverty=a;+a;'  +  a;'  +  a;*  .  .  . 

Ans,  x—y—y*+f^x*'\- 

2.  Reverty=a;— Ja;'  +  ia;'— la;*  .... 

Arts.  a;=y+iy'  +  iy'  +  |/ 

3.  Revert  y=a;—ia;Hia;^—|a;'  .... 

Ans.  x=y  +  if  +  ^jy'  +  ^\\f 

!•  Revert  y = 2a; +  3ic'  + 4a;' 4- 5a;'  .... 

Ans.  x^^JZ-T^y'  +  j'^y'-^WTf 

5.  Revert  y=l+a;+|a;'+J-a;'' 4- aV^*  ... 

Ans.  x=y-l-±{y-iy  +  ^{y-lY^^tZ^  .  . 

((•  Revert  y=a:4-2a;*4-4a;^  .... 

Ans.  a;=y— 2y'— 4y' 

7.  Revert  y=aJ—^aj^-f-ia;'—J-a5*  +  ia;'  ... 

4ns,  x=y^t  +  ^  +JL4.^L^  .  .  . 
^2      2-3     2-3-4      2-3-4-5 


«    -^  a;".       x'  x' 

8.  Revert  ^=0.--+-^:^^-^:^:^:^:^  .  .  . 

x'  ,  l-3a;'  ,    l-3-5a;' 
^^2-3     2-4-5      2-4-6-7^ 

9,  Reverty=:a;  +  3a;''  +  5a;'  +  7a;*  +  9a;' 

Ans.  a;=y— 3y'  +  13y''— 53/ 

*-.    T>  a;'      arV    a:*  ,    a;' 

10.  Reverty=a;+— +—  +  —+—  .... 

Z  O  'k  O 

Ans.  x=y--—-\-  —  —-^— 
^      2      2-3     2-8-4 


SUMMATION  OF  SEEIES.  416 

SUMMATION   OF    SERIES. 
The  summation  of  a  series  depends  upon  its  nature. 

ARITHMETICAL    SERIES. 
(426.)  The  summation  of  a  finite  arithmetical  series,  whether  in- 
creasing or  decreasing,  we  have  ah-eady  seen  is  embraced  in  the  for- 
mula ^=(— ^-Jw,  in  which  a  is  the  first  term,  I  the  last  term,  and 
n  the  number  of  terms. 

aEOMBTRICAL    SERIES. 
The  summation  of  a  finite  geometrical  series,  whether  increasing  or 

1.1/.        1     rt     ir—a        a—lr 
decreasmg,  is  embraced  m  the  formula  S= ~  or      _    . 

When  the  series  is  infinite  and  decreasing,  the  formula  is  fS=        -. 

RECURRING  SERIES. 

A  recurring  series  is  one  in  which  a  certain  number  of  consecutive 
terms,  taken  in  any  part  of  the  series,  has  a  fixed  relation  to  the  term 
which  immediately  follows. 

Thus,  in  the  series  l-{-3x  +  4:X*  +  1x'-{-llx*  +  18x''  ....  the  sum 
of  the  coeflScients  of  any  two  consecutive  terms  is  equal  to  the  coefl^- 
cient  of  the  following  term. 

In  the  series  l-^2x+3x^  +  4x^  +  5x*-\-Qx^ each  term 

after  the  second  is  equal  to  the  product  of  2x  into  the  first  preceding 
term  plus  the  product  of  —x^  into  the  second  preceding  term. 

Thus  5x^=2xx4x^—x^x3x'^.  The  coeflScients  of  x  and  a;',  or 
2—1,  is  called  the  scale  of  relation. 

Intheseriesl+4a;  +  6a;'-f  lla;'+28.r*4-63a:' ,2  — 14-3  is 

the  scale  of  relation.     Thus  63aj'— 2a:  x  28a:' —a;'  x  11a:'  +  3a:'  x  6x^. 

(427.)  The  fraction  -^ produces  the  series 

^  ^  0  +CX  ^ 

a      OCX      ac^x^      ac^x* 

h    W^'~Y      w 

ex 
in  which  each  term  after  the  first  is  equal  to  the  product  of  — j-  into 


44§  SUMMATION  OF  SEEIES. 

the  term  that  precedes  it.     In  this  series,  — —  is  the  scale  of  relation 

f> 
of  the  terms,  and  -  is  the  scale  of  relations  of  the  coefficients. 

CL  -I-  l)X 

(428.)  The  fraction  - — ; r  produces  the  series. 

^  ^  c  +  dx  +ex^  ^ 


a     Ih     ad\       /bd      ad^      ae\ 


in  which  each  term,  commencing  at  the  third,  is  equal  to  the  two 

cc        dx 

immediately  preceding  multiplied  respectively  by  — '—, ,  which 

c  c 

is  the  scale  of  relation  of  the  terms. 

ft  _L  J/j«  _1_  /»'>•' 

(429.)  The  fraction -— -.  will  produce  a  series  in 

d  +  ex  -irfx^-\-gx^  ^ 

which  each  term  commencing  at  the  fourth  is  equal  to  the  three  pre- 

yy/V*  "foi^  6X 

ceding  terms  multiplied  respectively  by  — ^,  —  ^^»  ~T>  ^^^^^  ^^ 
therefore  the  scale  of  relation  of  the  terms. 

(430.)  The  fraction „ '-^- t^t;  will  produce  a 

q-r-rx  +  sx^  .  .  .  ux''-\-vx''-^^  ^ 

series  in  which  each  term  commencing  at  the  n  +  2th  will  be  equal 

to  the  n  +  \  preceding  terms  multiplied  respectively  by 

vx"^^        u:f  sx^        rx 

T'  ~T'  ■  ■  •  •  •  ~'V  ~1' 

which  is  the  scale  of  relation  of  the  terms. 

When  ^=1  the  scale  of  relation  is  —  vaf+\  —war",  .  .  .  —sx^^  —rx, 
which  are  the  terms  of  the  denominator  taken  in  reverse  order  and 
with  contrary  signs,  the  first  term  being  omitted. 

(431.)  A  recurring  series  is  said  to  be  of  the  first  order  when 
each  term  commencing  with  the  second  depends  upon  the  one  that 
precedes  it. 

(432.)  A  recurring  series  is  said  to  be  of  the  second  order  when 
each  term  commencing  with  the  third  depends  on  the  two  preceding 
terms. 

In  like  manner  we  have  recurring  series  of  the  third  order,  fcmrth 
order,  &c. 

PiJOBLEM 
(433.)  1.  To  find  the  scale  of  relation  in  a  recurring  series  of  the 
first  order. 


SUMMATION  OF  SERIES.  417 

SOLUTION. 

Let  A,  B,  Cy  2),  &c.,  ...  be  the  coefficients  of  a  recurring  series 
of  the  first  order. 

Now,  some  relation  exists  between  each  two  consecutive  terms. 
Let  r  be  the  relation,  and  we  have  B—rA\  0=rB  ;  D—rC^  &c. 

whence,     r=— -=-^=:-;^,  &c. 


PROBLEM 
(434.)  2.  To  find  the  scale  of  relation  in  a  recurring  series  of  the 
second  order. 

SOL  UTION. 
Let  A,  B,  (7,  D,  &c.,  be  the  coefficients  of  the  series.    Whence  by 
the  conditions  must  arise  the  following  equations. 

C=zmA-\-nB, 
D=mB+nC, 
&c.,         &(5., 
in  which  m  and  n  are  unknown  quantities.     The  solution  of  these  two 
equations  must  give  their  values. 

BC=mAB-\-nB\ 

AB—mAB  +  nAC, 

AB-BC=(AC-B')n, 

_AD-BC 

""- AC~B'' 

In  the  same  way  m  may  be  found,  whose  value  is  seen  in  the  equation 

C'-BB 


m=z 


AC-B'' 


PROBLEM 
(435t)  3.  To  find  the  scale  of  relation  in  a  recurring  series  of  the 
third  order. 

SOL  UTION. 
Let  A,  B,  (7,  i).  By  F^  &c.,  be  the  coefficients.     By  the  conditions, 
we  must  have  B = mA  -\-nB-{-q  C, 

E=mB-\-nC+qB, 
F=mC-\-nB  +  qB, 
&c.,  &c. 

The  solution  of  these  equations  gives  the  values  of  m,  n,  and  q, 

27 


418  SUMMATION  OF  SERIES^ 

PROBLEM 
(436.)  4.  To  find  whether  the  law  of  the  series  depends  on  one, 
two,  three,  or  more  terms. 

SOLUTION 

See  whether  the  scale  of  relation  established  by  problem  1st  agrees 
with  the  given  series ;  if  not,  try  problem  2d ;  and  if  that  does  not 
agree,  try  problem  3d,  and  so  on  until  the  scale  is  found.  Or  we 
assume  that  the  series  is  of  a  higher  order  than  necessary,  in  which 
case  one  or  more  of  the  values  found  by  resolving  the  resulting 
equations  will  be  found  to  be  zero,  and  the  remaining  one  will  give 
the  true  scale  of  relation. 

PROBLEM 
(437,)  5.  To  find  the  sum  of  a  recurring  series  of  the  first  order. 

SOLUTION. 

It  is  evident  that  when  the  series  is  of  the  first  order  (that  is,  a 
geometrical  series),  the  sum  of  the  series  may  be  obtained  from  the 

formulas,  S= -,  S=- ,  and  S=- . 

The  first  formula  must  be  used  when  the  series  is  increasing  and 
finite ;  the  second,  when  it  is  decreasing  and  finite ;  and  the  third, 
when  it  is  decreasing  and  infinite. 

PROBLEM 

(438.)  6.  To  find  the  sum  of  an  infinite  recurring  series  of  the 
second  order. 

SOLUTION. 
Let  -4  + J?+  (7+2>  ....  be  a  series  of  the  second  order. 
Then  C=mAx''+nBx 
D=mBx'-^nCx 


Ii=mPx^-\-nQx  • 

Adding  these  equations  together,  and  putting  A  +  JB-h  C-\-D 
H  .  .  .  .^S,  we  shall  have 


SUMMATION   OF  SERIES.  419 

{\—nx—mx'^)S=A-{-B—nAx 
A  +  B—nAx 


S=- 


\—nx—mx* 


PROBLEM 

(439.)  *1.  To  find  the  sum  of  an  infinite  recurring  series  of  the 
third  order, 

SOLUTION. 

'  Let  A  +  B+  C+JD,  &c.  be  the  series. 

According  to  the  nature  of  a  recurring  series  of  the  third  order 
there  must  result  the  following  equation  : 

D = mAx^  +  nBx^  -\-qCx 

E=mBx^-\-nCx''-\-qDx 


Adding  these  equations,  and  putting  S=A-^B-\-  C+D-\-E,  &c., 

we  have    S—A—B—C=mSx'+n{S—A)x^-\-q{S—A'-B)x 

^A  +  B+C—{A-\-B)qx—nAx* 

wnence.  o — ~ —  „         , 

1—qx—nx—mx 

PROBLEM 

(440.)  8.  To  find  the  sum  of  a  given  number  of  terms  of  a  re- 
curring series  of  the  second  order. 

SOLUTION. 

Let  A-^B+  C+J)  ....  4-i2  be  a  finite  recurring  series  of  the 
second  order.    We  have 

C=mAx'  +  nBx 
J)=mBx^+nCx 


T=mBx^-{-nSx. 
Let  us  find  the  sum  of  the  recurring  series  of  the  second  order, 

U+  V+  W,  &c.,  to  infinity. 
We  have,  according  to  Prob.  6, 

o  =— 5- 

1—nx—mx 


420  SUMMATION  OF  SERIES. 

Supposing  A  +  JB+Cj  (fee,  to  be  an  infinite  series,  we  have  already 

found  its  sum  to  be 

A  +  B—nAx 

1—nx—mx^ 

Subtracting  from  this  the  sura  of  U-h  V+  TT,  &c.,  to  infinity,  and 

tbeie  must  remain  the  sum  of  the  finite  series,  A-]-£+0 .  .  .  .  T. 

Putting  this  sum  equal  to  S,  we  have 

^     A  +  B—U—V-nAx+nUx 

l—nx — mx 

Note. — ^In  the  same  way  we  might  obtain  the  sum  of  a  finite  re- 
cuning  series  of  higher  orders. 

PROBLEM 
(441,)  9.  To  find  the  general  term  of  a  recurring  series. 

SOLUTION. 

«^  ,  .       -       .       a-}-bx-\-cx*  .  .  .  . paf*  , 

The  general  generatmg  fraction,  ^_^_^^^^^.^    _  _^^^+v  may  be 

thus  represented : 

(a  +  bx  +  cx''  ,  .  .  .px'')(q-{-rx  +  sx'*  ....  +vx'^^)-^ 

The  second  parenthesis  may  be  expanded  by  the  multinomial 
theorem^  and  the  result  multiplied  by  the  quantity  in  the  first  pa- 
renthesis. Then,  if  we  take  in  this  product  the  part  which  contains  x 
to  any  power  whatsoever,  we  shall  obtain  the  general  term  of  the 
recurring  series. 

We  present  another  method. 

Take  the  generating  fraction, 

a-\-bx+cx^  .  .  , psf" 
q-{-rx+sx'^ .  .  .  vrr''+* 
which  is  supposed  to  be  reduced  to  its  lowest  terms. 

Dividing  both  numerator  and  denominator  by  v,  and  changing  the 
order  of  their  terms,  we  have 


V 

c  „     b       a 

.  .  .  -x^-\.-x-h- 

V            V          V 

a;"+\  . . 

s  „     r       a 

V            V          V 

Separate  the  denominator  into  binomial  factors.    Then  the  fraction 
may  be  considered  as  made  up  of  fractious  which  have  these  binomial 


SUMMATION   OF  SERIES.  421 

fractions  for  their  denominators.  Determine  these  partial  fractions 
according  to  the  method  given  in  undetermined  coeflScients,  and  then 
find  the  general  term  of  development  of  each  of  these  partial  fractions, 
and  the  sum  of  these  general  terms  will  be  the  general  term  of  the 
recurring  series. 

Note. — In  the  decomposition  into  partial  fractions,  when  a  factor 
of  the  denominator  is  of  the  form  (x-{-a){x-{-b)"'^  assume  the  fraction 
to  be  equal : 

A  B  C  JI_ 

x  +  a     {x-^-b)"^     {x  +  hy*-'  '  '  *  x  +  i) 
Each  partial  fraction  may  be  put  under  the  form  P^p+x)''  in 
which  r  is  some  positive  integral  number. 

The  expansion  of  this  by  the  binomial  theorem  gives  for  the  term 
which  contains  x""  the  following 

-r{-r-l)(-r~2)  ....  (-r-(«-l) 

r-2-3     .  .  .  .  «  ^^      "^  • 

It  is  the  sum  of  the  terms  containing  a:"  resulting  from  the  different 
partial  fractions,  that  composes  the  general  term  in  the  recurring 
series. 

(442.)  A  simpler  mode  of  finding  the  general  term  of  a  recurring 
series  will  be  found  in  the  following  discussion. 

We  have  already  seen  that  a  recurring  series  of  the  first  order  is 
a  geometrical  series,  and,  therefore,  its  general  term  is  easily  found. 

Let  us  suppose  that  a  +  ar  +  ar^  +  ar^  +  ar*  .  .  . ,  a  geometrical 
series,  is  also  a  recurring  series  of  the  second  order. 

When  this  is  the  case,  we  must  then  have  the  following  equations : 
ar^=:ma'\-nar, 
ar^=mar-\-nar^, 
ar* = mar^  +  war', 


ar*: 

in  which  m,  n  is  the  scale  of  relation.    - 

By  division  each  of  the  above  equations  reduces  to 

n±V^m-\-n^      x 

whence  r = . 

2 

If  these  two  values  of  r  are  not  equal,  we  see  that  there  may  be 
two  geometrical  series  which  will  also  be  recurring  series  of  the  second 
order,  having  m,  n  for  their  scale  of  relation. 


422  SUMMATION  OF  SERIES. 

Let  the  two  values  of  r  be  c,  and  h.  Then  since  the  first  terms  of 
"these  two  geometrical  recurring  series  of  the  second  order,  may  be 
any  quantities  whatever ;  let  the  variables  x  and  y  represent  them.   We 

shall  then  have 

x+xc  +  xc^-\-xc^  ......  xc'^\ 

yj^yh+yh'+yh' yh^^-K 

Each  of  these  series  are  recurring  and  of  the  secohd  order,  and  the 
scale  of  relation  in  each  is  m,  n. 

By  adding  them,  we  have 
{x-\-y)  +  {cx  +  hy)  +  (c''x-\-b^y)+{c'x+b'y) {xc^'+yh-% 

This  series  is  not  geometrical,  although  formed  by  the  addition  of 
two  series  which  are  geometrical.  But  these  two  series  are  also  re- 
curring series  of  the  second  order.  Have  we  a  right  then  to  conclude 
because  two  geometrical  series  added  together,  do  not  produce  a 
geometrical  series,  that  these  two  geometrical  recurring  series  have 
not  when  added  produced  a  recurring  series  ? 

Let  us  examine.     If  the  series 

{x  +  y)  +  {cx  +  hy)-\-{fx-\-h^y)  .... 

is  a  recurring  series  of  the  second  order,  whose  scale  of  relation  is  m, 
n,  we  must  have 

c^x  ■\-h'^y=m{x  +  y)-\-  n(cx  +  by) 
SinceaJ  +  caj-f-c'ic+c^  &c.,  and  y-\-by  +  ¥y-\-b''y,  &c.,  are  recur- 
ring series  whose  scale  of  relation  is  m,  %,  we  have  the  following 
equations :  c'^a; = mx  +  ncx 

b^yz=my  +  7iby. 
By  addition  c'x  +  b^y=m(x-^y)-\-n{cx-\-by)  which  proves  that 
{x  +  y)-\-{ax+by)  +  {c'x+b'y)  ....  {xa^-'+yb^') 

is  a  recurring  series  of  the  second  order,  whose  scale  of  relation  is  m,  n. 
Let  this  series  be  represented  by 

A-^B+C-^I)  ......  T;  7"  standing  for  the 

generaL  term.    The  following  equations  must  then  subsist : 

x-^y=A, 
cx  +  by=:Bf 

B-bA 

'  whence,  x= — , 

'  c—b 

B-cA 


2^="-^:=6- 


We  also  have  T  =:a?c^'  +  ^6" 


^    B-bA     ,     B-cA' 

T=- — j-C^' j-6*-*. 

c—b  c—o 


SUMMATION  OF  SERIES,  423 

By  assuming  a  geometrical  series  to  be  a  recurring  series  of  the 
third  order,  we  should  find  r  by  solving  a  cubic  equation.  Then  pro- 
ceeding in  a  similar  manner,  we  should  find  that  a  recurring  series  of 
the  third  order  is  equal  to  the  sum  of  three  geometrical  series,  having 
the  same  scale  of  relation,  whence  the  general  term  for  a  series  of  the 
third  order  might  be  obtained. 

• 

EXAMPLE  S 

1.  Find  the  sum  of  1  +  2a:  +  Sa;"  4-  28a;'  +  100a;'  +  356a;',<feo. 

1— a; 


Ans,  Scale  2a;\  3a; ;  sum 


'       '  l-3a;— 2a;' 

2,  Find  the  sum  of  1  4-  2a;  +  3a;'  +  4a;'  +  5a;*,  &c. 

Ans.  Scale  —a;,  2a; ;  sum  - 


1  — 2a;  +  a;' 

3.  Find  the  sum  of  l+3a;  +  4a;'  +  7a;''  +  lla;'  +  18a;'  +  29a;',  &c. 

Atis.  Scale  a;",  x\  sum -v. 

1 — X—TT 

4.  Find  the  sum  of  1 +a;+5a;''  + 13a;' +  41a;*  +  121a;' +  365a;«  &c. 

J /p 

Am.  Scale  Sa;*,  2a; ;  sum  - — ~r. 

'      '  1— 2a;— 3a?' 

5*  Find  the  sum  of  1  +  6a;  + 1 2a;»  +  48a;'  + 1 20a;*,  <fec. 

l+Sa; 


Ans,  Scale  Ga;",  x\  sum 

f  5a;' +  •?«?'  + 9a;*  4-1  la;*,  <S 
Ans,  Scale  —  a;\  2a; ;  sum 


1-a;— 6a;»* 
6.  Find  the  sum  of  1 4-  3a;  4-  5a;'  4-  "^aj'  4-  9a;*  4- 1  la;*,  &c. 

14-.'P 


1— 2a;-|-a;' 
?•  Fmd  the  sum  of  1 4- 3a;  4- 2a:'— a;'— 3a;* --2a;' 4- a;',  <fec. 

l  +  2a; 


Am,  Scale  —a;',  a; ;  sum 

«    -r,.    1  .1  ^  a     ac      ac*  ,     «c'  ,    . 

8.  Fmd  the  sum  of  T"~TT^  +  T?a;  — ti^\  &c. 

o       0  0  0 


1— a;4-a;'* 


Ans.  Scale  —  ^a?;  sum 


h   '  64-caJ 

9,  Find  the  sum  of  3  4-  6a; 4-  7a;'  4-  13a;'  4-  23a;*,  &c. 

.«        CI    1         «  a     a   «  3— a;— 6a;' 

Ans.  Scale  —2a;',  a;",  2a; ;  sum r r. 

'     '       '  1  — 2a;— a;''4-2a;» 

10.  Find  the  sum  of  1 4-a?  4-  2a;'  4-  2a;'  4-  3a;*  4-  3a;*  4-  4a;'  4-  4a;',  &c. 

Ans.  Scale  —a;',  a;',  xi  sum z -,. 

1— a-— a;'4-a;' 


424  SUMMATION  OF  SERIES. 

11.  Find  the  sum  of  1 +4a;  +  6a;'  +  lla;''  +  28.r*  +  63a;^  <fec. 

Ans.  Scale  Sx^,  —x^,  2x:  sum  \- (^ — — -«• 

'         '       '  {l—xY—3x^ 

12.  Find  the  wth  term  of  1+ 2a;  +  3a;' +  4a;' +  5a:*,  &c. 

Ans.  nx*^^, 

13.  Find  the  wth  term  of  1  +  3a;  4-  Bx""  +  7a;'  +  9x\  <fec. 

Ans.  (2w— l)af-\ 

14.  Find  the  wth  term  of  l+a;  +  5a;'  +  13a;''4-41a;*  +  121a;^&c. 

Ans.  ±(3a;)"-^  +  i(-a;)'^*. 

15.  Find  the  wth  term  of  1 +a;  +  3a;''  +  5a;'  +  lla;*4-21a;^  &c. 

Ans.  |(2a;)"-'  +  i(— a;)**-*. 

16.  Find  the  sum  of  n  terms  of  1  +  2a;  +  3a;' 

.       1  — (/i4-l)a;'*  +  waf*+* 

Ans.  ^^ , 

.1  — 2a;  +  a;' 

17.  Find  the  sum  of  n  terms  of  1  + 3a;  +  5a;' ^-Ya;' 

ns.  ______  . 

18.  Find  two  geometrical  recurring  series,  which  when  added  shall 
be  equal  to  the  series  l+a;+5a;' f  13a;'  +  41a;*  .  .  . 

C  i  +  lia;  +  4ia;'  +  13ia;'  +  40ia;*  .  .  . 

19.  Find  two  geometrical  recurring  series  whose  sum  shall  be  equal 
tothe  series  l+a!  +  3«'  +  5a:'  +  lla;*  .... 


U-k^  +  i^-i-'  +  i^'  .  .  . 


20.  Find  three  geometrical  recurring  series  whose  sum  shall  be  equal 
totheseries  l+a;  +  2a;'  +  3a;'  +  6.r'  .... 

(i  +  i^  +  K  +  K  +  K 

Ans.    U^^x  +  ^x'  +  ix'-h\^x* 

(i-i^  +  K-i^'+i^' 

(443.)  An  increasing  series  is  one  whose  successive  terms  in- 
crease in  value. 

(444.)  A  decreasing  series  is  one  whose  successive  terms  decrease 
in  value. 

(445..)  A  converging  series  is  one  in  which  the  greater  the  num- 
ber of  terms  taken,  the  nearer  will  their  sum  approximate  the  true 
sum  of  the  series. 


SUMMATION  OP  SERIES.  425 

PROBLEM. 
(446.)  To  find  the  degree  of  approximation  when  a  limited  num- 
ber of  terms  are  taken  in  a  decreasing  converging  series  whose  terms 
are  alternately  positive  and  negative. 

SOLUTION. 
Leta— 6  +  c— df+ -\-m"-n+p—q-\-r— rep- 
resent a  decreasing  converging  series,  whose  odd  terms  are  positive, 
and  even  terms  necrative. 

o 

Since  a  is  greater  than  b,  c  than  J,  m  than  w,  it  follows  that  a— -6, 
c—d,m—n,p—q^  &c.,  are  positive  quantities. 

CASE     I. 

When  the  number  of  terms  taken  is  even. 

Leta— 6+c— c?+ m—w+^—g' be  the  series  to  n  terms, 

n  being  even. 

Since  r— «,  l—u,  and  the  other  rejected  couplets  are  each  positive, 
it  follows  that  the  above  series  to  the  n  terms,  when  n  is  even,  is  less 
than  the  true  sum  of  the  whole  series.  Hence,  we  have  the  following 
inequation,  a— 6  + c—c? m—ri+^—g'<>S,  in  which  ^rep- 
resents the  true  sum  of  the  whole  series. 

Now  let  a—h-\-c—d m—n-^p—q-\-r  be  the  series  to 

n-\-\  terms,  n-\-\  being  an  odd  number. 

The  terms  after  r  are  —s  +  t—u  +  v These  terms  may 

be  thus  represented,  —  (s— <)—(%— v)—  ....  The  quantities  in 
the  parenthesis  are  positive,  and  since  each  parenthesis  is  negative,  it 
follows  that  the  series  which  terminates  in  r  is  followed  by  a  series  of 
negative  quantities,  whence  the  following  inequation,  a—h-\-c^d 
m—n-\-p—q-\-r'^S. 

From  this,  we  see  that  the  quantity  necessary  to  be  added  to  the 
first  member  of  the  first  inequation  to  make  it  equal  to  S  must  be  less 
than  r. 

CASE    II. 
When  the  number  of  terms  taken  is  odd. 

Let  a—h-\-c—d m— w+^  be  the  series  to  n  terms,  n 

being  odd. 

Reasoning,  as  in  the  last  case,  we  have  the  following  inequations, 

a—h  +  c—d m—n+p'^S, 

and     a — h  +  c — d  .....  m — n-^p—q<^S^ 


426  THE  DIFFERENTIAL   METHOD. 

which  show  that  the  quantity  to  be  ^btracted  from  the  first  member 
of  the  first  inequation  to  make  it  equal  to  *S^  must  be  less  than  q. 
The  solution  of  this  problem  establishes  the  following 

THEOREM. 
(4:4:T,)  The  error  committed  by  taking  n  terms  of  a  decreasing 
converging  series^  whose  terms  are  alternately  positive  and  negative^ 
for  the  sum  of  the  whole  series,  is  numerically  less  than  the  following 
term. 

Note. — In  applying  this  theorem  to  certain  examples,  the  series 
should  be  considered  as  commencing  at  the  second  term  of  the  ex- 
pansion. 


i^  »>  ♦  •>  »■ 


THE   DIFFERENTIAL    METHOD. 

(448.)  The  differential  method  of  summing  a  series  is  a  method 
of  finding  the  sum  of  a  series  by  ascertaining  the  successive  differences 
of  its  terms. 

PROBLEM. 
(449.)  1.  To  find  the  first  term  of  any  order  of  differences. 

SOLUTION. 

Let  a,  6,  c,  d,  e,f  <fec.,  be  a  series. 

Supposing  this  series  to  be  ascending,  we  have  a  new  series  by 
taking  the  first  term  from  the  second,  the  second  from  the  third,  and 
go  on.    This  series  is  called  theirs/  order  of  differences. 

In  like  manner  we  may  take  the  differences  of  the  successive  terms 
of  this  new  series,  and  thus  form  still  another  series,  which  is  called 
the  second  order  of  differences.  By  continuing  the  process,  we  should 
obtain  other  orders  of  differences.     Thus,  we  may  have 

1st  order  b— a,  c—b,  d—c,  e—d,f—e 

2d  order        c—2b-\-a,d—2c-{-b,e—2d-hc,f—2e-\-d  . 

3d  order  d~Sc  +  3b—a,e—3d  +  Sc—b,f—Se-\-3d—c  .  .  .  . 

4th  order  e— 4c?  +  6c— 46  +  a,/— 4e  +  6c?— 4c-f-6 

6th  order  f^5e  +  10d—l0c  +  5b—a 

&c.,  <fec.,  (fee. 


THE   DIFFERENTIAL   METHOD.  '42^ 

By  inspecting  the  coeflScients  of  the  terms  in  the  successive  order 
of  differences,  we  see  that  in  the  first  order  they  are  the  same  as  the 
coeflScients  of  the  expansion  of  (x—yY  ;  in  the  second  order,  the  same 
as  the  coeflScients  of  the  expansion  of  {x—yY  ;  in  the  third  order,  as 
those  in  the  expansion  of  {x—yY  ;  and  in  the  n\h  order,  the  coeflS- 
cients must  be  the  same  as  those  in  the  expansion  of  (ar— y)".  The 
first  in  each  of  the  above  order  of  differences  may  be  written  thus ; 

1st  order  —{a—h). 

2d  order  a—2b  +  c. 

3d  order  —(a—db-\-3c—d). 

4th  order  a—4b-i-6c—4:d  +  e. 

5th  order  —  (a— 56  +  lOc— 10c?  +  5e— /). 
For  the  wth  order,  we  see  that  we  must  have  the  following  formula, 
in  which  J}^  stands    for    the  wth    order  of   differences,   D^=db 

^  2  2.3  2.3.4  ^ 

The  upper  sign  before  the  parenthesis  must  be  used  when  n  is  even, 
and  the  lower  one  when  n  is  odd. 

PROBLEM 
(450.)  2.  To  find  the  nth  term  of  a  series. 

SOLUTION. 

By  Prob.  1.  (449)  we  have  the  following  equations^ 

i>,=6— a, 

i>2=c-26  +  a, 

B^^d—'Sc  +  Sb—a, 

i>4=:e— 4c?  +  6c— 46  +  a, 

i>5=r/~5e  +  10(]?— 10<;+66— a, 

&c.,  &c. 

Whence  may  be  obtained 
b=a  +  D^, 
c=:a  +  22)^+2>2, 

&c.,  &c. 

We  see  that  in  the  {n-\-\)i\i  term  of  the  series  a,  6,  c,  c?,  e,  <fec.,  the 
coeflScients  of  a,  -Z),,  D^^  J)^,  D^,  &c.,  are  the  same  as  the  coeflS-  ' 
cients  in  the  expansion  of  (x  +  y)".     Therefore,  the  {n-\-l  )th  term  of  the 


series 

Z>4  +,  &c.,  from  whicli  we  see  tiiat  the  nth  term 
2      •      3     .     4 


428  THE  DIFFERENTIAL  METHOD 

,        ,    «       .  ^       7i(w»  — 1)^       w(w— l)(n— 2)^ 

a,  6,  c,  d,  <fec.,  IS  a+w2>,  +  -^—^^IB^  +-^^ — ^'^-^-^s  + 

«(w— l)(7i  —  2)(?i— 3) 
2      ;      3     .     4 

must  bea  +  (7i-l)l>,+-^^ ^^^ -■Di  +  - — ^r-^"^ —  ^^^ 

It  2     .      o 

(w-l)(n-2)(7i--3)(w-4)  _,      ,  ..  1.  .      vx  •     J  V 

4-^-— — ^^^-- — '^         '^ I^^t  <»c.,  which  is  obtained  by  writing 

n— 1  for  n, 

PROBLEM 
(451*)  3.  To  find  the  sum  of  n  terms  of  a  series. 

SOT^UTION. 

Let  a,  6,  c,  <?,  <fec.,  be  the  proposed  series,  of  which  we  seek  to  find 
the  sum  of  n  of  its  terms. 

Taking  the  series  0,  a,  a + 6,  a  +  6  +  c,  a  +  6  +  c  +  <?,  <fec.,  we  see  that 
its  (w  +  l)th  term  is  equal  to  the  sum  of  n  terms  of  the  series 
a,  6,  c,  c?,  &c. 

Representing  the  first  term  of  each  order  of  difierences  in  the  series 

0,  a^a-^h^  a  +  6+c,  &c.,  respectively  by  7>,,  D^^  -Dg,  D^^  &c.,  we 

have  by  the  last  problem  for  the  (w  +  l)th  term  of  the  series  0,  o, 

a  +  6,  a-\-h-\-c^  a-\-h-\-c-\-d^  &c.,  or  the  sum  of  n  terms  of  the  series 

,     «  -^        ^(^— l)-r^        7i(«  — l)(w  — 2)  ^     . 

a,   6,   c,   (^,    &c.,  -0  +  nB,  +    ^   ^     ^A  +    \     %     ^  A  + 

^^_l)(^_2)(^-3) 

2.3.4^*  +,  <fcc. 

If  we  wish  to  refer  the  first  term  of  each  order  of  difierences  to  the 
proposed  series  a,  6,  c,  c?,  <fec.,  instead  of  to  the  assumed  series  0,  a, 
a^-5^a_}-5  +  c,  a  +  ft  +  c  +  c?,  &c.,  the  formula  becomes  (since  7>,,  J^g* 
Z>3,  &c.,  of  the  assumed  series  are  respectively  equal  to  a,  Z>i,  D^^  &c. 

of  the  proposed  senes)  8=na^—^-- — -D^^ — ^-- — '-^^ — ^-^a  + 
2.3.4  ^'       ' 

EXAMPLES. 

1,  What  is  the  first  term  of  the  fourth  order  of  difierences  of  the 
series  1,  8,  27,  64,  125,  &c.?  Ans.  0. 

2.  What  is  the  first  term  of  the  eighth  order  of  difierences  of  the 
series  1,  3,  9,  27,  81,  &c.  ?  Ans,  256. 


THE  DIFFEKENTIAL  METHOD.  429 

3.  What  is  the  first  term  of  the  fifth  order  of  diflferences  of  the 
series  1,  i,  i,  },  j\,  3V,  eV,  &c.  ?  Ans,  -  3V 

4.  What  is  the  first  term  of  the  sixth  order  of  differences  of  the 
series,  3,  6,  11,  17,  24,  36,  50,  72,  &c.?  Ans.  —14. 

5.  What  is  the  first  term  of  the  third  order  of  differences  of  the 
series  1,  2*,  3*,  4*,  &cJ.  Am.  60. 

6.  What  is  the  10th  term  of  the  series  1,  4,  8,  13,  19,  &c. « 

Ans.  64. 

7.  What  is  the  15th  term  of  the  series  1',  2\  3\  4^  &c.  ? 

Ans.  225. 

8»  What  is  the  20th  term  of  the  series  1,  3,  6,  7,  &c.  ? 

Ans.  39. 

9.  What  is  the  wth  term  of  the  series  «,  a+c?,  a+2c?,  a  +  3c?,  <fec.! 

Ans.  a-{-(n—l)d. 

10.  What  is  the  nih.  term  of  the  series  1,  3,  6,  10,  15,  21,  &c.  ? 

Am.  "("  +  '\ 
2 

11,  What  is  the  50th  term  of  the  series  1,  4,  8,  13,  &c.  ? 

Ans,  1324. 

12»  What  is  the  sum  of  n  terms  of  the  series  1,  3,  5,  7,  9,  &c.  ? 

Ans.  n*, 

13*  What  is  the  smn  of  n  terms  of  the  series  1',  2",  3^*,  4",  <fec.  ? 

n{n  +  l){2n  +  l) 
'1.2.3 
lit  What  is  the  smn  of  n  terms  of  the  series  1,  2,  3,  4,  5,  &c.  ? 

Ans.  !<!^). 
2 

15*  What  is  the  sum  of  n  terms  of  the  series  1',  2",  3",  4^  &c.  ? 

Ans.  }{n*-\-2n^-{-n^). 

16.  What  is  the  sum  of  n  terms  of  the  series  1,  2*,  3*,  4*,  &g.  ? 

.         n^  ,   n*      n^      n 

Ans. , 

6        2       3      30 

17.  What  is  the  sum  of  »  terms  of  the  series  1,  8,  27,  64,  <fec.? 

An,.  ^>±1)\ 


480        APPLICATION  OF  THE  DIFPEHENTIAL  METHOD. 

18.  What  is  the  sum  of  n  terms  of  the  series  1,  3,  6,  10,  15,  &o.  1 

7i(%+l)(w  +  2) 

Ans,  -\ ^  ■  ;^    '  , 

1.2.3 

19.  What  is  the  sum  of  n  terms  of  the  series  l(w  +  1),  2(w+2), 

8(m4-3),  4(m  +  4),  &c*?  Ans,     '^    ;  o ^-^—^- 

1     •     2     .     3 


APPLICATION    OF    THE    DIFFERENTIAL    METHOD. 

PROBLEM. 
(452.)  To  find  the  number  of  balls  or  shells  in  a  triangular  pile. 

SOLUTION. 

By  an  inspection  of  a  triangular  pile  of  balls,  or  shells, 
we  see  that  the  first  tier,  commencing  at  the  top,  has  one 
ball ;  the  second,  1+  2  ;  the  third,  1  +  2  +  3 ;  the  4th, 
1  +  2  +  3  +  4;  and  the  nth,  1+2  +  3+4  +  5  .  .  .  +w. 
The  number  of  tiers  equals  the  number  of  balls,  or  shells, 
in  one  side  of  the  base. 

Then,  to  find  the  number  of  balls,  or  shells,  in  a  triangular  pile  of 

n  tiers,  we  must  find  the  sum  of  the  series  1,  3,  6,  10, ....  to  » 

terms,  which  is 

_n{n+\){n  +  2) 

1.2.3 

PROBLEM. 
(453.)  To  find  the  number  of  balls  or  shells  in  a  square  pile. 

SOLUTION. 

By  an  inspection  of  a  square  pile  of  balls, 
or  shells,  we  see  that  the  first  tier,  com- 
mencing at  the  top,  has  I'' ;  the  second,  2^* ; 
the  third,  3^^ ;  the  fourth,  4^* ;  and  the  nth, 
w'  balls,  or  shells. 

Then,  to  find  the  number  of  balls,  or  shells,  in  a  square  pile  of  n 
tiers,  we  must  find  the  sum  of  the  series  l^  2',  3",  4',  5',  to  n  terms, 
which  is  n{n^l)(inj^) 

1.2.3 


APPLICATION  OF  THE  DIFFERENTIAL  METHOD.        431 

PROBLEM. 
(45  4«)  To  find  the  number  of  balls  or  shells,  in  an  oblong  pile, 

SOLUTION. 

By  an  inspection  of  an  oblong  pile  of  balls,  or  shells,  we  see  that  if 
m  + 1  represents  the  number  in 
the  first  tier,  commencing  at  the 
top,  the  number  in  the  second  is 
2(w+2);  intheSd,  3(w^  +  3)• 
in  the  4th,  4(m  +  4) ;  and  in  the 
nth,  n(m  +  n). 

Then,  to  find  the  number  of  balls,  or  shells,  in  an  oblong  pile  of  n 
tiers,  we  must  find  the  sum  of  l(m  +  I),  2(//i4-2),  3(m  +  3), , . . .  to  n 
terms,  which  is 

w(%  +  l)(l+2w4-3m) 

1.2.3 
(455«)  These  three  formulas  may  be  written  as  follows  : 
In  the  triangular  pile,  3=^  .  -^^— — •  .  (w  + 1  + 1) 

"    "     square        "      5=i .  ^^^^\  (w  +  w  +  l) 

"     "     oblong       **      5=:i.^?i!!±D  i(ri+m)  +  (w+m)+(m+l)} 

Since  -~- — -   represents  the  number  of  balls,  or  shells,  in  the 
ii 

triangular  face  of  each  pile,  and  the  other  factor  the  number  in  the 

longest  base  line,  plus  the  number  in  the  line  parallel  to  it,  plus  the 

number  in  the  top  tier,  we  have  the  following 

GENERAL     RULE. 

Multiply  one-third  the  number  in  the  triangular  face  hy  the  num' 
her  in  the  longest  base  line^  increased  by  the  number  in  the  opposite 
parallel  line  and  by  the  number  in  the  top  tier,  and  the  product  will 
be  the  whole  number  of  balls  in  the  pile, 

EXAMPLES. 

1*  How  many  balls  in  a  triangular  pile  of  16  courses  ? 

Ans,  680* 


432  SPECIAL  SERIES. 

2.  A  complete  square  pile  of  shells  has  14  courses :  how  many 
shells  are  in  the  pile,  and  how  many  remain  after  the  removal  of  5 
courses  ?  Ans.  1015  in  the  pile,  and  960  remain. 

3*  In  an  incomplete  oblong  pile  of  balls,  the  length  and  breadth  at 
the  bottom  are  respectively  46  and  20,  and  the  length  and  breadth  at 
the  top  are  35  and  9.     How  many  does  it  contain  ?       Ans.  7190. 

4.  The  number  of  balls  in  an  incomplete  square  pile  is  equal  to  6 
times  the  number  removed,  and  the  number  of  courses  left  is  equal  to 
the  number  of  courses  taken  away.  How  many  balls  were  in  the 
complete  pile  ?  Ans.  385. 

5*  Let  h  and  k  denote  the  length  and  breadth  at  the  top  of  an 
oblong  truncated  pile,  and  n  the  number  of  balls  in  each  of  the  slant- 
ing edges.    How  many  in  this  truncated  pile  ? 

Ans.  ^J2n^  +  3n{h+k)  +  6hk—S{k-[-k+n)+l^, 

6«  How  many  shot  in  a  triangular  pile  whose  bottom  row  contains 
8  shot?         '  Ans.  120. 

7.  How  many  shot  in  a  square  pile  whose  bottom  row  contains  8 
shot?  Ans.  204. 

8.  How  many  shot  in  an  oblong  pile  whose  length  and  breadth  at 
the  bottom  contains  respectively  16  and  T  ?  Ans.  392. 

9.  How  many  shot  in  a  triangular  pile  whose  bottom  row  contains 
30  shot  ?  Ans.  4960. 

10.  How  many  shot  in  a  square  pile  whose  bottom  row  contains  30 
shot-?  Ans.  9455. 

1 1 .  How  many  shot  in  an  oblong  pile  whose  number  of  courses  is 
30,  and  top  row  31  ?  Ans.  23405. 

12.  How  many  shot  in  an  incomplete  oblong  pile  whose  length 
and  breadth  at  the  bottom  are  respectively  46  and  20,  and  at  the  top 
85  and  9?  •    Ans,  1190, 


i»  ■■  »  t.  »■ 


SPECIAL  SERIES. 
THEOREM. 


1 

(456.)  Any  fraction  of  the  form       ^        is  equal  to  -  of  tha 


q  q 

difference  between  the  fractions  -  and  . 


SPECIAL  SEKIES.  488 

DEMONSTBATION. 

g q    ^qn-hpg-nq^      pq      ^  q      _l/g_    9    \ 

n     n+p        n(n+p)        n(n-\-p)^'   '  n{n-\-p)    p\n     n-\-p)' 

Q.  K  D. 

OoROLLABY. — If  the  difference  between  -  and  — —  is  Imown,  the 
•^  n         n+p 

value  of  —r^ — c  will  be  known,  whether  -  and  — —  are  known  or 
n(n  +p)  n         n  -\-p 

not. 

Hence,  we  obtain  the  following 

PRINCIPLE. 

In  any  series  of  fractions  having  the  form  — ^-^ — r  the  sum  of  the 

n{n-\-p)  ^ 

series  is  equal  to  -th  of  the  difference  between  a  series  of  fractions  of 
the  form  -,  and  a  series  of  fractions  of  the  form 


QUESTION . 

(45Ti)  What  is  the  sum  of  the  series  t:^+-^  +  —  +...  to  in- 
finity? 

SOUTTION. 


Here 


q 


n{n-{-p)     1*2 

q    -  1 

n{n-\-p)     2*3 
q      _  1 


•.  §'=1 ;  n=\  ;  w+jt>=2,  or^=l. 
'.  ^=1;  71=2;  w+^=3,  or^=l. 
•.  3'=1;  w=3;  w+^=4,  or^=l. 


n{n  4-j9)     3*4 

From  which,  we  see  that  q  constantly  equals  1  and  p  constantly 
equals  1,  and  n  successively  equals  1,  2,  3,  <feo. 

Therefore,  ^=^+^+1  +  ^. ..  to  infinity, 
w     1     2     3     4  ^ 

n     n+p 


Whence,  -(--— ^|,  or  -  of  1,  which  =1  is  the  sum  of  a  series  of 
p\n    n+p/       p 

ictiona 

infinity. 


fractions  of  the  form  -— — r,  or  of  the  series  r-i  +  r-i  +--•  +  ...  to 
n(n+py  1-2     2-3     3*4 


28 


434  SPECIAL   SERIES. 

Note. — ^In  the  following  examples  some  difficulties  are  purposely  left  for  the 
student  to  overcome.  But  it  is  believed  that  the  skill  that  he  has  acquired  by 
mastering  the  work  thus  far  will  enable  him  to  solve  them. 

QUES  TION  S. 

1,  What  is  the  sum  of  the  series  — -  4-—  +  -^^  + to  infinity? 

X'o      ^'4      o'O 


2,  "What  is  the  sum  of  the  series  t:^~^  +^k~"  •  •  *^  infinity  ? 

I'D        ^*4       0*0 


Ans.  |. 

infinity  ? 
Ans.  A. 


4        4        4  4 

3«  What  is  the  sum  of  the  series  t-l  +3^  +:rT7.  +  :: — tt;  +  ...  to 

1-5     5-9     9-13     13-17 

infinity  ?  .  Ans,  1, 

4,  What  is  the  sum  of  the  series  -—■  +r-i  +r-7,+Tii  + to 

».     1*4     2*5     3*o     4*7 

infinity?  Ans.  {}. 

5i  What  is  the  sum  of  the  series  7-7:  +;r-L +^-w  + to  infinity? 

1*3      3'5      5*7 

Ans.  |. 

6.  What  is  the  sum  of  n  terms  of  the  series  :ri+r-^  +  ;r^  +  TH  "^ 

1*4      2*6      3*o      4*7 

<kc?  Ans.  +-; — r  + 


3w  +  3     6W4-12     9w  +  27 
7.  What  is  the  sum  of  n  terms  of  the  series  7-^+^:^  ■^;^+»<fe' 

I'o       o'O      O*  I 

Ans. 


2w  +  l 


8.  What  is  the  sum  of  the  series  ^-i""7-K+Hi^"~7rT-i  "*"  •  •  •  *o  "^" 

3*5      5*7      7*9      y*li 

finity?  Ans.  yV- 

9.  What  is  the  sum  of  the  series  1  +-—+-+--+,  <fec.,  to  infinity! 

8       o      iU 

Ans.  2. 

10.  What  is  the  sum  of  the  series --+^^+^7Y^+,&c.,  to  infinity! 

Ans.  ■^. 


SPECIAL  SERIES.  ,435 

1 1 ,  What  is  the  sum  of  n  terms  of  the  series  —  +  — -  +  — -  + ,  &c.? 

12.  What  is  the  sum  of  n  terms  of  the  series  oT^  "~  ^  "^  ^T^  —  oTTi  ''" » 

4o.?  Ans.  jf^dcz—z r,  accordinff  as  n  is  odd  or  even. 

^^      4(2w  +  3)  ° 

By  reasoning  as  in  (456),  we  obtain  the  following 

PRINCIPLE. 

(45  8  i)  If  any  series  of  fractions  have  the  form— ^ \/  _i-9  V 

the  sum  of  the  series  is  equal  to  —  of  the  difference  between  a  series 

of  fractimis  of  the  form  —r-^ — -.  and  a  series  of  fractions  of  the  form 
9 


{n-^p){n  +  2p)' 

QUESTIONS. 

12  3 

1 ,  What  is  the  sum  of  the  series  ^-r—  +  ^ttt;  "^  Tir?:  +  >  ^^'^  *^  ^^' 

I'o'o       o'o'i      5*7'9 

finity.  Ans,  |. 

2.  What  is  the  sum  of  the  series  -±-+^+ J^  +  -^+, 

&c.,  to  infinity  ?  Ans.  -^j. 

3  9  15 

3*  What  is  the  sum  of  the  series  — — --+  (..-..,.-,.+  n.iAo^  "^^ 

<fec.,  to  infinity  ?  Ans.  ^A* 

A  K  fi 

4.  What  is  the  sum  of  the  series  +  +  + ,  &c.,  to 

1*2*3      2*3*4       3*4*0 

infinity  1  Ans.  1^. 

5.  What  is  the  sum  of  the  serie? 

a  a  +  b  a  +  26 

n(n-\-p){n-\-2p)  "^  {n-\-p){n-{-2p)(n-{-Sp)      {n-{-2p)(n  +  Sp){n  +  4i^y 

.   „  .     „  .  pa  +  bn 

&c.,  to  mfinity  ?  Ans.   ^  ,   . — — ^. 

'  "^  2jo'^w(%+jp) 


436  SPECIAL  SERIES. 

Again,  by  reasoning  as  in  (456),  we  obtain  the  following 

PRINCIPLE. 

(459*)  If  any  series  of  fractions  have  the  forn^ 


1.  What  is  the  sum  of  the  series  ^^^^  +;r7r7T7'^;r77^:^+»<^'» 


n(n  -\-p)  (n  +  2p)n  +  3^)' 
the  sum  of  the  series  is  equal  to  —  of  the  difference  between  a 

of  fractions  of  the  form  —. %-, -—-^.  and  a  series  of  fracti/ms  of 

''  ''  J        J         w(7i+jp)(w  +  2^y  ''  ''  '' 

the  form  —, r-. — - — 77 -— r. 

•^        n{n+p){n  +  2p){n  +  Zp)     * 

QUESTIONS. 

l-2-3'4     2-3-4-5     3-4-5-6 
to  infinity  ?  Ans.  y^. 

12  3 

8,  What  is  the  sum  of  the  series  — --  +  ^^^^  +  5:7:9:1 1  +» ^^-^ 

to  infinity  ?  Ans.  y^. 

2                   5                     8 
3a  What  is  the  sum  of  the  series + + + , 

9.  vvii<ttiBiu«Buuiuitu«»«ic  3.6.9-12     6-9-12-15     9-12-15-18     ' 

<fec.,  to  infinity  ?  Ans.  gf  |s . 

/»a  /72  qs 

4.  What  k  the  sum  of  the  series  ,^-^^^  + ^^-+ ^^^^+ , 

&c.,  to  infinity  ?  Ans.  |f . 

Again  reasoning  as  in  (456),  we  obtain  the  following 

PRINCIPLE. 

(460i)  In  any  series  of  fractions  of  the  form 

? 

n{n+p){n-\-2p)  ....  (w+mp)  ' 

the  sum  of  the  series  is  equ>al  to  the of  the  difference  between  a 

''  ^  mp 


series  of  fractions  of  the  form  —.         .-7       ~"^  /       ,        i\  \' 

•^  -^  J        J         n{n^p)(n-\-2p)  ....  [n  +  {m—l)pj 

and  a  series  of  fractions  of  thefwm  -. tt — — —  r 7 — ; ^. 

Again  reasoning  as  in  (456),  we  obtain  the  following 


SPECIAL  SERIES.  487 


PRINCIPLE. 

(46  It)  In  any  series  of  fractions  of  the  form 
a(a-{-b)  ....  (a+pb) 
n(n  +  b)  .  .  .  ,  (n+pby 

the  sum  of  the  series  is  equal  to of  the  difference  between  a 

j:  J-     *'         j^  *-L    X        '  «(«  +  &)  ....  {a-\-pb) 

series  of  fractions  of  the  form    —7 r^ f ^7-^ — tt^*  «^»  a 

''  '^  J         J  n{n  +  b)  ....  [n-h{p—l)by 

series  of  fractions  of  the  form  —} y!  '  '  '  '  ; it- — —^* 

•^  "^  ^        ^  n(n  +  b)  ....  (n+pb) 


QUESTIONS. 

1.  What  is  the  sum  of  r  terms  of  the  series -  +  ——  +  — —-  + 

1-3-5-7  ^     0     ,  .       l'3-6-'7  .  .   .  .  (2r+l)     ^ 

— — — — +  ,&o.?  Ans. — — ^ ^— 1. 

2-4-6-8      *  2-4-6  ....  27* 

2     2*4      2'4'6 

2.  What  is  the  smn  of  r  terms  of  the  series  -4--—+  + 

0     Z'o      0*0*7 

2-4-6-8  ^    .     ,  .        2-4-6-8  .  .  .  .  (2r+2) 

?5W  +  '*"-'  ^'"-  3-5-T-9  ■  ■  ■  ■   (2r +!)-'• 

9  2*3         2*3*4 

3.  What  is  the  smn  of  the  series  rrr  +  ^-^77:  +  t^-^ttt  + »  <^c*»  ^ 

0*0      o*D'7     o*D*7*8 

infinity?  Ans.  ^. 

4*  What  IS  the   sum  of  r   terms  ot  the  senes  -  +  •  ' 


n     n(n  +  b) 
a(a  +  b){a  +  2b)  ^ 

n(n  +  b)(n-h2b)  ^^'^'"^ 

^^^    1       /  a(a  +  b){a+2b) (a  +  rb)      \ 

'  n—a-b\       n{n+b){n  +  2b)  ....  [n  +  {r—l)b])' 

Remark. — If  7i=a  + 26,  whence  ni-b=a  +  Sb  and  n  +  2b=a-{-4by 
&c.,  and  n  +  (r—2)b=a  +  rb,  the  fraction  in  the  parenthesis  becomes 

a(a  +  b) 
a  +  {l^r)b' 
which  vanishes  when  r  is  infinite,  and  we  therefore  have  for  the  sum 
of  the  series  to  infinity 

a 
n—a^h' 


438  SPECIAL  SERIES. 

When  71= a  4- ft,  the  fraction  in  the  parenthesis  becomes  a,  and  the 

sum  of  the  series -(a--a)=-(0)=-,  an  expression  of  no  defi- 

n — a — 0  0  0 

nite  signification  in  its  present  form. 

It  may  be  observed  that  this  is  obtained  without  reference  to  the 

value  of  r ;  hence,  whether  r  is  infinite  or  finite,  the  result  is  the  same. 

(46!2#)  Some  series  may  be  very  beautifully  summed  as  follows. 

QUESTION 

1.  What  is  the  sum  of  the  infinite  series  x  +  x*-}-x'-\-x*-\-y  <fec.  ? 


SOLUTION. 

Let     S=x+x''-\-x^-hx*+x^-h,&;G. 
then    xS=   -{■x'  +  x^  +  x*-\-x^  +  ,&;c. 
.-.     {l-x)S=x. 

whence     S = . 

1—x 

QUESTION 

2.  What  is  the  sum  of  the  infinite  series  a;— rc'  +  ir'— aj*+,  &c.  ? 

SOLUTION. 

Let     S=x—x^-\-x^—x*-{-y&;c. 
then     —xS=—x''+x''—x*  +  y&c. 


•.    {l+x)S.=x. 
whence,     S=: 


1-\'X 


QUESTION 

3.  What  is  the  sum  of  the  infinite  series  x-{-2x^  +  3x*  +  4x*+y  &qJ 

SOLUTION. 

Let     S=.x  +  2a;'  +  3a;'  +  4a;* +,  <fec. 
then     —2xS=   -2a;'— 4a;'— 6a;*-,  <fec. 
and    x^S=  +  a;'  +  2a;*H-,  <fcc. 

.*.     (1— 2a;4-a;')=a;. 


SPECIAL  SERIES.  439 

QUESTION 

4.  What  is  the  sum  of  the  infinite  series  a;  4-  4a;'  +  9x'  +  16a;*  + ,  &c.  ? 

SOLUTION. 

Let     S=x  4-  4x^  +   9x^  + 1 6a;*  + ,  &c. 

then     —SxS=—Sar'  —  12x'  —  2lx*—,&c, 

and     Sx''S=  +   Sx^ -t  I2x*  + ,  Sua. 

and     —x^S—  —     x*—,  <fcc. 

.-.    (l'-Bx  +  Sx''-x')S=x  +  x\ 

x(l-^x) 


whence,     S: 


(i-xy 


PROMISCUOUS      QUESTIONS. 

!•  What  is  the  sum  of  the  infinite  series  H-a;  +  a;+.T'+  <fec.1 

1 


Ans. 


1—x 


2.  What  is  the  sum  of  the  infinite  series  1  +  2a;  +  3a;'  +  4a;'  +  5x*  + , 

1 


<fec.  ?  Ans. 


(i-^r 


3.  What  is  the  sum  of  the  infinite  series  T:7r^+};:^-:  +  ^^rr  +  >  &c.  ? 

1*^*3      ^*o*4      o'4'o 

3  4  5 

4*  What  is  the  sum  of  the  infinite  series  — —  +  +         3  + ,  (fee.  ? 

1*2*2     2*3*2       0*4*2 

Ans.  1. 
tj  ft 

5*  What  is  the  sum  of  the  infinite   series  ,:o.Q.oa  +  o.Q.^.oa  + 

1*2*3*2       2*3*4*2 

^+,&c?  Ans.i. 

6.  What  is  the  sum  of  the  infinite  series  +in.or  +  ToTo"^~  + 

8*18       10*21       12*24 

1  3 

+  ,<fec.?  Ans. 


14*27      '  80 

-.       .rrr,  .         ,  ^      ,  .     ^     .  .  10*18  12*21 

7t  What  is  the  sum  of  the  innnite  series 1 1- 

■  •    TTxm.,  10  luic  Duixi  w*  i,ixc    xiiiixiL^  2*4*9*12      4*6*12*15 

14*24  ^    ^  .19 

+  ,&c.?  Ans. 


6*8*15*18      '  54 


440  SPECIAL  SERIES. 

8.  What  is  the  sum  of  the  infinite  series  x  +  Sx^i-  6a;'  +  10a;*  + ,  &c.  ? 


Ans. 


9.  "What  is  the  sum  of  n  terms  of  the  series ■  h , 

4-8      6-10  ^8-12      ' 

&c.  ?  Ans. 


16(1  +  n)     12(3  +  2ny 

4         4  4  4 

10.  What  is  the  sum  of  the  infinite  series  — —  -\-  -— -  +  — —  -f  + 

I'o      5*9      9*13      13*17 

+  ,  <fec.  Ans,  1. 


17*21 


THE    END, 


J 


